Do black holes rip apart even atoms and protons and neutrons? I am aware that black holes rip apart objects because of spaghettification. What about atoms and protons and neutrons? Do they get ripped apart too? But, in the case of protons and neutrons, wouldn't them getting ripped apart into quarks violate the fact that you can't isolate quarks?
 A: The "spaghettification force" you're referring to is better known as the tidal force on the object.  To within an order of magnitude, it is proportional to
$$
F_\text{tidal} = \frac{G m M d}{r^3}
$$
where $m$ is the mass of the object, $d$ is its physical size, $r$ is the radial coordinate, and $M$ is the mass of the black hole.
For an object just crossing the event horizon of a black hole of radius $r$, we have $M = c^2 r/2G$, and so this becomes
$$
F_\text{tidal} = \frac{m c^2 d}{2 r^2}
$$
For a black hole of radius $r = 3$ km and an atom ($d = 10^{-10}$ m, $m = 10^{-27}$ kg), this force works out to be about $10^{-34}$ N.  Hardly anything to worry about;  for comparison, the force between hydrogen nucleus and its electron is about $10^{-8}$ N.
Of course, as the atom gets closer to the singularity $r \to 0$, this force will increase.  But you'd have to make $r$ very small to get that to work.  In such regimes, it would not surprise me if well-known phenomena of quantum field theory in normal regimes (in particular, color confinement) broke down.  But, of course, such situations can never be observed from the outside universe.
A: You have another nice answer from  Michael Seifert which explicitly computes the tidal force on a hydrogen atom at the event horizon of a solar-mass black hole. That force is much smaller than the electric attraction which binds the electron and proton together, so hydrogen is apparently not "spaghettified" on the outside of a solar-mass black hole.
The tidal stretching is larger outside of smaller black holes, varying like $1/r^2 \propto 1/M^2$.
A calculation you might like to try would be to compute the radius-or-mass of a black hole whose tides are strong enough to dissociate hydrogen at the event horizon. Likewise, there is some less-massive black hole whose exterior tides are strong enough to separate nucleons from a nucleus, and an even-less-massive black hole whose tides are strong enough to generate meson excitations from baryons like protons or neutrons.
This handy calculator (hat tip to PM 2Ring) computes the tidal acceleration $\mathrm d\kappa_R$ in $\rm (m/s^2)/m$.
For two objects with equal mass $m$ separated by a distance $d$, the tidal force pulling each away from the center, near the Schwartzchild radius $R$, will be
\begin{align}
F_\text{tidal} &= m(a - a_\text{center}) 
\\ &= m\left(\mathrm d\kappa _R \cdot \frac d2\right)
\\ &= m \frac{c^2}{R^2} \frac d2
\end{align}
which is the same as the expression in Michael's answer.
For microscopic systems we usually don't talk about forces and accelerations, but instead about energies and distances.
Let's abuse our notation and use $U_\text{tidal} = F_\text{tidal} d$ as an estimate of the "tidal energy," and say that a system get dissociated when the tidal energy is bigger than the binding energy. This is probably wrong by some factors of two, which is typical for dimensional-analysis estimates.  Setting $U_\text{tidal} = U_\text{binding}$ gives
$$
R = \sqrt{\frac{mc^2 d^2}{2 U_\text{binding}}}
= d \sqrt{\frac{mc^2}{2 U_\text{binding}}}
$$
as the radius of a black hole whose tidal force at the event horizon can dissociate a system of size $d$ with binding energy $U_\text{binding}$ and constituent mass $m$.
Some order-of-magnitude results, all using the proton mass of $mc^2 = 1\,\rm GeV$:




system
length $d$
energy $U$
hole radius $R$
hole temperature
hole mass




electron from atom
$10^{-10}$ m
~10 eV
$10^{-6}$ m
200 K
$10^{20}$ kg


nucleon from nucleus
$10^{-15}$ m
~10 MeV
$10^{-14}$ m
$10^{10}$ K
$10^{12}$ kg


meson from nucleon
$10^{-15}$ m
~200 MeV
$10^{-15}$ m
$10^{11}$ K
$10^{11}$ kg




These black holes are all microscopic in size, but
I don't know whether these qualify as quantum-regime black holes. The masses are many orders of magnitude above the Planck scale, where weirdness is guaranteed; the temperatures involved are high, but comparable to the temperatures where these same phenomena happen in accelerators. Nevertheless, treat these estimates with some suspicion.
A: The sphagettificaton force isn't infinite, it is of order of magnitude $GM/r^{2}$.  This can get quite large for small black holes${}^{1}$, but there will be some upper limit to the effects it can have.
${}^{1}$ at the schwarzschild horizon, $r=2GM/c^{2}$, so we have the "sphagettification" force being roughly $ GMc^{4}/(4G^{2}M^{2}) = c^{4}/4GM$, so for a stellar mass black hole, this is  comes out to like $10^{13} N$, which is certainly very large, but not infinite.
