Since you've already received a good answer to the full question, I'll just add a bit more detail on the analogue of the Aharanov-Bohm effect for gravity. As noted in the existing answer and elsewhere, any analogy between gravity and gauge theory can't be perfect because the graviton has spin $2$.
On the other hand, we don't need a perfect analogy if we just want an analogue of the Aharanov-Bohm effect, because the standard setups for it (and its relatives, such as the Aharanov-Casher effect) all involve experimental equipment that breaks Lorentz invariance. After all, you would never analyze this effect in a frame where the solenoid was moving! If we don't bother to maintain manifest Lorentz invariance, we are free to butcher gravity until it takes the shape we want.
First analogy: Newtonian limit
In the standard Newtonian limit of general relativity, the metric reduces to
$$ds^2 = - (1 - 2 \Phi) \, dt^2 + (1 + 2 \Phi) \delta_{ij} \, dx^i dx^j$$
where $\Phi$ is the Newtonian potential, and is sourced by matter in the standard way,
$$\Phi(x) = \int d \mathbf{x}' \, \frac{\rho(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}.$$
The action for a particle is
$$S = - m \int \sqrt{-g_{\mu\nu} u^\mu u^\nu} \, d\tau$$
where $u^\mu$ is the four-velocity. In the nonrelativistic limit $u^\mu \approx (1, \mathbf{v})$, this becomes
$$S \approx - m \int \sqrt{1 - 2 \Phi - (1 + 2 \Phi) v^2} \, dt \approx \int \frac{mv^2}{2} (1 + 2 \Phi) - m(1+\Phi) \, dt.$$
Neglecting the $\Phi v^2$ term as small, since both $\Phi$ and $v^2$ are, this is a standard "kinetic minus potential" Lagrangian, where the mass-energy $m$ and the gravitational potential energy $m \Phi$ are counted in the potential. Therefore, an object picks up an extra gravitational phase shift of
$$\Delta \phi = - \frac{m}{\hbar} \int \Phi \, dt.$$
This phase shift is commonly measured by atom interferometry.
Is this the gravitational Aharanov-Bohm effect? According to this paper, it is. After all, it's a phase that can be measured by interferometry, which accumulates even when the gravitational field at the particles themselves vanishes. On the other hand, the phase shift doesn't have the $\mathbf{A} \cdot \mathbf{v}$ form, so it depends on the rate at which we traverse a path, while the usual Aharanov-Bohm effect doesn't. It doesn't have the same "geometric" flavor.
Second analogy: Gravitoelectromagnetism
To get something closer, we can back up and do a less crude approximation. In the weak field approximation to general relativity, where $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$, the field $h_{\mu\nu}$ is sourced by matter as
$$\bar{h}_{\mu\nu}(\mathbf{x}) = 4 \int d\mathbf{x}' \, \frac{T_{\mu\nu}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}$$
where we have defined the trace reversal $\bar{h}_{\mu\nu} = h_{\mu\nu} - \eta_{\mu\nu} h / 2$ and set $c = G = 1$. Now, this looks quite similar to how the vector potential is sourced by a current,
$$A_\mu(\mathbf{x}) = \int d\mathbf{x}' \, \frac{J_{\mu}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}$$
but we have an extra tensor index. Now we can abandon Lorentz invariance to go further. Note that for nonrelativistic source matter with typical speed $u$, the stress energy tensor components are
$$T_{00} \sim O(1), \quad T_{0i} \sim T_{i0} \sim O(u), \quad T_{ij} \sim O(u^2).$$
So in the nonrelativistic limit, we could choose to neglect the $T_{ij}$ (and hence the $\bar{h}_{ij}$) entirely. Then, since the stress-energy tensor is symmetric, $T_{i0}$ is redundant with $T_{0i}$, so we only need to keep track of the elements $T_{0\mu}$. This is the same number of degrees of freedom as a four-vector.
By expanding $\bar{h}_{ij}$ to leading order in $u$, we get the slightly more general metric
$$ds^2 = - (1 - 2 \Phi) \, dt^2 + 2 (\mathbf{A} \cdot d \mathbf{x}) \, dt + ( 1 + 2 \Phi) \, \delta_{ij} dx^i dx^j$$
where $A_\mu = (\Phi, \mathbf{A})$ is the gravitoelectromagnetic (GEM) four-potential, and
$$A_0 = \Phi = \int d\mathbf{x}' \, \frac{T_{00}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}, \quad A_i = \int d\mathbf{x}' \, \frac{T_{0i}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}.$$
That is, $T_{0\mu}$ sources $A_\mu$ just like $J_\mu$ sources $A_\mu$ in electromagnetism. Of course, the analogy is not perfect because $T_{0\mu}$ isn't a four-vector, so our $A_\mu$ doesn't have nice Lorentz transformation properties, but nonrelativistic source matter already picks a preferred frame anyway.
For a slowly moving test particle, a similar analysis to the Newtonian limit gives
$$S \approx \int \frac{mv^2}{2} - m(1+\Phi) + m \mathbf{v} \cdot \mathbf{A} \, dt$$
where we are working to quadratic order in $u$ and the speed $v$ of the test particle. The coupling to $\mathbf{A}$ looks just like the coupling to the magnetic vector potential, which implies that the force on the particle obeys the usual Lorentz force law, but with charge $m$ and "gravito-electric" and "gravito-magnetic" fields, which are defined in terms of $\mathbf{A}$ just as in electromagnetism. For more on this idea, which is used to analyze precision relativity experiments such as Gravity Probe B, see here.
Finally, the phase shift due to encircling a gravito-magnetic flux is
$$\Delta \phi = \frac{m}{\hbar} \int \mathbf{v} \cdot \mathbf{A} \, dt = \frac{m}{\hbar} \int \mathbf{A} \cdot d \mathbf{x} = \frac{m \Phi_B}{\hbar}.$$
Since $\mathbf{E}$ and $\mathbf{B}$ are defined analogously to electromagnetism, $\mathbf{B}$ obeys the Biot-Savart law with current density $T_{0i}$, which is the momentum density. That is, the gravitational analogue of a solenoid is a rotating cylinder. In a more geometric language, where we hadn't sliced away the Lorentz invariance, this would be described as a "frame dragging" effect. (Explicitly, the magnetic flux through a cylindrical solenoid is $\mu_0 A J$ where $A$ is the area and $J$ is the current density. The computation of the gravitomagnetic flux is identical, except that $J$ is replaced with $\sigma v$, the product of the surface mass density and velocity of the cylinder.)
So, is this the gravitational Aharanov-Bohm effect? Again, the analogy is decent, but not perfect. In the nonrelativistic limit, the phase shift indeed looks identical to that of the ordinary Aharanov-Bohm effect. And the gravitomagnetic field does vanish everywhere outside the cylinder. But there's one catch: the gravitoelectric field (i.e. the ordinary gravitational field) doesn't vanish outside the cylinder, while the electric field outside a solenoid does. This deficiency is because there are no negative gravitational charges, so our gravitational solenoid is like a regular solenoid if the electrons were rotating without any compensating protons. As with everything else, it can ultimately be traced back to the fact that the graviton has spin $2$ instead of $1$, since even-spin mediated forces are universally attractive. (For what it's worth, this ultimately comes back to bite electromagnetism as well: the fact that charges can cancel out, leaving only currents, implies that it has two distinct nonrelativistic limits.)
In summary, while there are no perfect analogies to the Aharanov-Bohm effect, there exist at least two approximate ones. The first is measured in real experiments; the second arises from a formalism used to analyze other real experiments. Would it someday be possible to measure the gravitomagnetic phase shift through atom interferometry? It seems very difficult now, since it is penalized by a factor $u$ of the source matter speed, but who knows in the future!