What happens to an atom when it gets sucked into a black hole? Lets say a helium atom is pulled into a black hole, does it stay a helium atom? What happens to it?
 A: The helium atom most certainly does not remain a helium atom, and on first thought, the answer is, "No,...duh!". On further inspection, the details of the $^4_2{\rm He}^0$'s journey to oblivion require more careful consideration.
For simplicity, consider a radial trajectory into a classical Schwarzschild black hole, qualitatively. Once the atom crosses the event horizon, all time-like trajectories point to the singularity, where (or when) the atom's constituents cease to exist.
On the free fall trajectory towards the singularity, the system is disrupted by tidal forces, or gravitational gradients. Classically, when
$$m_eg(r+a_0) - m_eg(r-a_0) 
\approx R_{\infty} $$
you would expect the atomic shells to be thoroughly disrupted. Here $g(r)$ is equivalent "local gravitational acceleration", aka "force of gravity"...which may not even be a thing (it diverges at the event horizon), but lets suppose there is some sort of analog along the trajectory.
$$a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_ee^2} \approx 50\,{\rm pm}$$
is the Bohr radius, and:
$$ R_{\infty} = \frac{m_ee^4}{4\epsilon_0^2\hbar^3c}\approx 13.8\,{\rm eV}$$
is the Rydberg constant.
Basically, your perfectly spherically symmetric $^4_2{\rm He}^0$ is going to have an induced quadruple moment that couples to the 'field' gradient, and at some point that coupling becomes larger than the binding energy so that the electrons are stripped. This is popularly called spaghettification.
Thus:
$$^4_2{\rm He}^0\rightarrow ^4_2{\rm He}^{2+} + 2e^- = \alpha + 2e^-$$
and the atom is officially destroyed.
The electrons are "point particles" (sort of), and proceed nicely towards the singularity. If the tidal forces disrupt the electrons' surrounding cloud of virtual particles is a possibility, but beyond the scope of this.
The alpha particle is not this:

Rather, it is a perfectly round collection of 4 nucleons. It is both scalar and iso-scalar, meaning a single nucleon is half spin up and half spin down (coherently), and half proton and half neutron (coherently). It is roughly 4 proton's in size:
$$ r_{\alpha} = 3.6\,{\rm} $$
with a binding energy of 28.3 MeV. Presumably it also becomes spaghettified via
tidal forces:
$$ \alpha \rightarrow 2n+2p $$
and the finally state nucleons will be entangled in a spin-0, $I=0$, state, though as a dynamical system, there should be virtual $\pi$, $\omega$, $\rho$, etc, doing their effective field theory thing.
If or how the protons are shredded into quarks via:
$$ p \rightarrow 2u + d $$
(and likewise for the neutrons) is difficult to say, as QCD is non-linear, and free quarks aren't a thing.
At this point, it's important to note that quarks and electrons aren't physical point particles, rather, they are excitations of the quark and electron fields, respectively. Physics breaks down at the singularity, but according to Kip Thorne, the 'particles' reach the singularity and cease to exist. Their mass is included in the total black hole mass, which exists in the curvature of spacetime. The black-hole is a stable soliton solution, made of entirely of space-time, around an inexplicable (for now) point singularity of infinite curvature. The helium atom is gone, the protons and neutrons are gone, the quarks and electrons are gone.
Whether information about them is also gone is major outstanding problem in physics, called the Black Hole Information Paradox.
A: I would distinguish two phases, when the atom crosses the horizon and when the atom reaches the singularity. When the atom crosses the event horizon of the black hole nothing happens to it, in fact, nothing special happens at the event horizon and you wouldn't even notice you've crossed it. After it crosses the horizon, the atom's fate is to reach the singularity, and at that point, nobody actually knows what happens.
