In the case of electrostatics, We know that for electric field $\mathbf{E}$, $$\nabla \times \mathbf{E}=0$$
which in the line with the integral form $$\oint \mathbf{E}\cdot d\mathbf{l}=0$$
Now Can I take this loop directly through the atom's nucleus? How do I know if it remains valid?
... because the electric field $\boldsymbol E$ is not defined at the nucleus. You can take the loop to pass as close as you want, so long as it does not touch it. The value $\boldsymbol E(0)$ doesn't exist, so you cannot evaluate the integral if it contains the point $r=0$, it is just a meaningless expression.
The reason is that $E\sim 1/r^2$, so the electric field is not defined at the origin. Formally it diverges, so the integral $\int_0 1/r^2$ doesn't exist. (In some situations the singularity can be mild enough for the integral to converge. But when this happens, convergence is most often conditional, so the result is ambiguous; it depends on the actual path you use to define the improper integral).
More precisely, the fields are defined only on $\mathbb R^3\setminus S$, where $S$ is the set of singularities of the sources. The PDEs that define $\boldsymbol E,\boldsymbol B$ (roughly, $\partial E\sim\rho$, $\partial B\sim j$) only make sense on $\mathbb R^3\setminus S$; the set $S$ is not part of the domain thereof. Point charges have $\rho\sim \delta(r)$, so the electric field is undefined at the origin.
The discussion above is in the context of classical electromagnetism, where the nucleus is assumed pointlike. If you want a more detailed analysis, where the nucleus has substructure, then the question becomes ill-defined unless you specify which substructure you want to work with.
For example, you may want to take the nucleus to be described by a charge density distribution $\rho(\vec x)$, which is peaked around $\vec x=0$ but varies continuously. If $\rho(\vec x)$ is finite everywhere, then the electric field will be finite everywhere too, and you can take your loop to be wherever you want. Of course, in this situation "through the nucleus" is kinda meaningless, because $\rho$ is extended so the nucleus is not at a point, but rather smeared out in space. But anyway, as long as $\rho$ is sufficiently well-behaved, the electric field will be smooth too, and you can take the loop to go through the region where $\rho$ is peaked.
On the other hand, if you want the nucleus to be described by, say, a collection of quarks, all of which are pointlike, then we are back to the initial situation: you can set the loop wherever you want except the location of the quarks. This description will not be better than the classical one so it is a rather pointless exercise, but it is self-consistent.
A better description would be taking into account that quarks are moving at relativistic speeds. But in this situation we would not be looking at static electromagnetism so again the question is meaningless.
An even better description would be using the fact that quarks are in fact quantum-mechanical, but again the situation is not static electromagnetism so again the question is meaningless.
In any case, the philosophy is the same as in the first part of this answer: you can take your loop to be anywhere you want as long as the electric field is defined there. This field is defined everywhere except for its singularities, i.e., the points where the sources $\rho,J$ diverge. This happens, for example, when the charges are pointlike so $\rho\sim\delta(r)$. So if your nucleus is pointlike, or is composed of pointlike objects, then the loop can be anywhere except for the location of these charges. If your nucleus is extended in space, and smooth everywhere, then the loop can be anywhere.
The two equations are (mathematically) equivalent only if the loop path can be continuously shrinked to a point in a way that preserves $\nabla \times \mathbf E = 0$ at all points of the loop.
This isn't so, for example, for a loop that is laced around infinite solenoid in which current changes (loop plane parallel to solenoid coils). This is because electric field outside of long solenoid obeys $\nabla\times \mathbf E =0$, but it does not obey this equation inside the solenoid.
Back to your question, we always can consider a loop that goes through the atomic nucleus, whether it is point or extended region of space. Then the loop integral may or may not exist and it may or may not be zero. All depends on details of the model of the nucleus.
In the simplest models, electric charge in nucleus is static. If in addition electric charge has finite charge density everywhere, the line integral exists. If charge density has singularity then it may not exist - it depends on details of the distribution of charge.
In a more realistic model, nucleus both moves with acceleration as a whole and its constituents have mutual accelerated motions, so electric charge is not static nor moving rectilinearly. We can imagine nucleus as kind of drop of very dense charged liquid. Then based on EM theory magnetic field of nucleus has to change in time and $\nabla \times \mathbf E$ may not vanish there. Then the integral $$ \oint \mathbf E \cdot d\mathbf l $$
may have various values, depending on the exact path chosen.
If the loop is large, it would work as the above comments indicate. If it is small, classical field theory would not work. Quantum effects are strong at that spatial scale: the location of the nucleus and the momentum, the orbital and spin of electrons around, beta-decay, the screening effects of vacuum turning into virtual electron-positron pairs, etc. It would probably only make sense to talk about the quantum fields involving photons as the carrier of the electromagnetic fields.
