In this specific instance you are correct. If you have a hydrogen atom that is completely isolated from the environment, and which has been prepared in a pure quantum state given by a superposition of the $1s$ and $2p$ states, then yes, the charge density of the electron (defined as the electron charge times the probability density, $e|\psi(\mathbf r)|^2$) will oscillate in time.
In essence, this is because the $2p$ wavefunction has two lobes with opposite sign, so adding it to the $1s$ blob will tend to shift it towards the positive-sign lobe of the $p$ peanut. However, the relative phase of the two evolves over time, so at some point the $p$ signs will switch over, and the $1s$ blob will be pushed in the other direction.
It's worth doing this in a bit more detail. The two wavefunctions in play are
$$
\psi_{100}(\mathbf r,t) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} e^{-iE_{100}t/\hbar}
$$
and
$$
\psi_{210}(\mathbf r, t) = \frac{1}{\sqrt{32\pi a_0^5}} \, z \, e^{-r/2a_0} e^{-iE_{210}t/\hbar},
$$
both normalized to unit norm. Here the two energies are different, with the energy difference
$$\Delta E = E_{210}-E_{100} = 10.2\mathrm{\: eV}=\hbar\omega = \frac{2\pi\,\hbar }{405.3\:\mathrm{as}}$$
giving a sub-femtosecond period. This means that the superposition wavefunction has a time dependence,
$$
\psi(\mathbf r,t)
= \frac{\psi_{100}(\mathbf r,t) + \psi_{210}(\mathbf r,t)}{\sqrt{2}}
=
\frac{1}{\sqrt{2\pi a_0^3}}
e^{-iE_{100}t/\hbar}
\left(
e^{-r/a_0}
+
e^{-i\omega t}
\frac{z}{a_0}
\frac{
e^{-r/2a_0}
}{
4\sqrt{2}
}
\right)
,
$$
and this goes directly into the oscillating density:
$$
|\psi(\mathbf r,t)|^2
=
\frac{1}{2\pi a_0^3}
\left[
e^{-2r/a_0}
+
\frac{z^2}{a_0^2}
\frac{
e^{-r/a_0}
}{
32
}
+
z
\cos(\omega t)
\,
\frac{e^{-3r/2a_0}}{2\sqrt{2}a_0}
\right]
.
$$
Taking a slice through the $x,z$ plane, this density looks as follows:
Mathematica source through Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["https://i.stack.imgur.com/KAbFl.png"]
This is indeed what a superposition state looks like, as a function of time, for an isolated hydrogen atom in a pure state.
On the other hand, a word of warning: the above statement simply states: "this is what the (square modulus of the) wavefunction looks like in this situation." Quantum mechanics strictly restricts itself to providing this quantity with physical meaning if you actually perform a high-resolution position measurements at different times, and compare the resulting probability distributions. (Alternatively, as done below, you might find some other interesting observable to probe this wavefunction, but the message is the same: you don't really get to talk about physical stuff until and unless you perform a projective measurement.)
This means that, even with the wavefunction above, quantum mechanics does not go as far as saying that "there is oscillating charge" in this situation. In fact, that is a counterfactual statement, since it implies knowledge of the position of the electron in the same atom at different times without a (state-destroying) measurement. Any such claims, tempting as they are, are strictly outside of the formal machinery and interpretations of quantum mechanics.
Also, and for clarity, this superposition state, like any hydrogen state with support in $n>1$ states, will eventually decay down to the ground state by emitting a photon. However, the lifetime of the $2p$ state is on the order of $1.5\:\mathrm{ns}$, so there's room for some four million oscillations of the superposition state before it really starts decaying.
A lot of atomic physics was forged in a time when a nanosecond was essentially instantaneous, and this informed a lot of our attitudes towards atomic superposition states. However, current technology makes subpicosecond resolution available with a modest effort, and femtosecond resolution (and better) is by now routine for many groups. The coherent dynamics of electrons in superposition states has been the name of the game for some time now.
It's also important to make an additional caveat: this is not the state that you will get if you initialize the atom in the excited $2p$ state and wait for it to decay until half of the population is in the ground state. In a full quantum mechanical treatment, you also need to consider the quantum mechanics of the radiation field, which you usually initialize in the vacuum, $|0⟩$, but that means that after half the population has decayed, the state of the system is
$$
|\Psi⟩= \frac{|1s⟩|\psi⟩+|2p⟩|0⟩}{\sqrt{2}},
$$
where $|\psi⟩$ is a state of the radiation field with a single photon in it, and which is therefore orthogonal to the EM vacuum $|0⟩$. What that means is that the atom and the radiation field are entangled, and that neither can be considered to even have a pure quantum state on its own. Instead, the state of the atom is fully described (for all experiments that do not involve looking at the radiation that's already been emitted) by the reduced density matrix obtained by tracing out the radiation field,
$$
\rho_\mathrm{atom}
= \operatorname{Tr}_\mathrm{EM}\mathopen{}\left(|\Psi⟩⟨\Psi|\right)\mathclose{}
=\frac{|1s⟩⟨1s|+|2p⟩⟨2p|}{2},
$$
and this does not show any oscillations in the charge density.
Foundational issues about interpretations aside, it's important to note that this is indeed a real, physical oscillation (of the wavefunction, at least), and that equivalent oscillations have indeed been observed experimentally.
Doing it for this hydrogen superposition is very challenging, because the period is blazingly fast, and it's currently just out of reach for the methods we have at the moment. (That's likely to change over the next five to ten years, though: we broke the attosecond precision barrier just last week.)
The landmark experiment in this regard, therefore, used a slightly slower superposition, with a tighter energy spacing. In particular, they used two different fine-structure states within the valence shell of the Kr+ ion, i.e. the states $4p_{3/2}^{-1}$ and $4p_{1/2}^{-1}$, which have the same $n$ and $L$, but with different spin-orbit alignments, giving different total angular momenta, and which are separated by
$$\Delta E=0.67\:\mathrm{eV}=2\pi\hbar/6.17\:\mathrm{fs}.$$
That experiment is reported in
Real-time observation of valence electron motion. E. Goulielmakis et al. Nature 466, 739 (2010).
They prepared the superposition by removing one of the $4p$ electrons of Kr using tunnel ionization, with a strong ~2-cycle pulse in the IR, which is plenty hard to get right. The crucial step, of course, is the measurement, which is a second ionization step, using a single, very short ($<150\:\mathrm{as}$) UV burst of light.
Here the superposition you're probing is slightly more complicated than the hydrogen wavefunction the OP asks about, but the essentials remain the same. Basically, the electron is in a superposition of an $l=1,m=0$ state, and an $l=1,m=1$ state, with an oscillation between them induced by the difference in energy given by the spin-orbit coupling.
This means that the shape of the ion's charge density is changing with time, and this will directly impact how easy it is for the UV pulse to ionize it again to form Kr2+. What you end up measuring is absorbance: if the UV ionizes the system, then it's absorbed more strongly.
The absorbtion data therefore shows a clear oscillation as a function of the delay between the two pulses:
The pictures below show a good indication of how the electron cloud moves over time. (That's actually the hole density w.r.t. the charge density of the neutral Kr atom, but it's all the same, really.) However, it's important to note that the pictures are obviously only theoretical reconstructions.
Anyways, there you have it: charge densities (defined as $e|\psi(\mathbf r)|^2$) do oscillate over time, for isolated atoms in pure superposition states.
Finally, the standard caveats apply: the oscillations caused in quantum mechanics by superpositions are only valid for pure, isolated states. If your system is entangled with the environment (or, as noted above, with the radiation it's already emitted), then this will degrade (and typically kill) any oscillations of local observables. If the overall state of the world is in some meaningful superposition of energy eigenstates, then that state will indeed evolve in time. However, for heavily entangled states, like thermal states or anything strongly coupled to the environment, any local observables will typically be stationary, because each half of an entangled state doesn't even have a proper state to call its own.
