Aren't Black Holes Just Neutron/Quark Stars? Disclaimer:
My understanding of general relativity is rudimentary at best, so bare with me and correct me where necessary. Also, any math in the explanation is appreciated, but try to also keep the answer intuitive for rookies like myself :)

The short version:
Aren't black holes just neutron\quark  stars compressed at least slightly beyond the associated Swartzschild radius?

The long version:
Per my understanding of general relativity, if I were to replace a star with an equally massive black hole, I should find that the space-time curvature from the boundary of where the star's surface used to be and beyond should stay the same, and I should also find that the space-time in the region between said boundary and the event horizon should continue to curve such that the curvature is continuous at the boundary. If I continue to probe\calculate the curvature up until the center of mass, I should find that it grows to infinity, but why should I be able to do so without running into any spatial boundary?
Consider neutron or quark stars stars (if the latter exist), where degeneracy pressure prevent the matter from compressing any further. It is also the case that if such star would cross a certain mass threshold, then the Swartzschild radius would become bigger than the radius of the star, thus resulting in a black hole.
So the intuition I get is that it would be fair to assume that when a core of a star is massive enough to collapse into a black hole, we should still expect to find all its mass compressed as much as the degeneracy pressure would allow it - we just wouldn't be able to observe it since that mass is enclosed within the event horizon. This would also imply that asking what is the space-time curvature at the center of mass (where the result is infinity) is meaningless, since the degenerate mass poses a spatial boundary, beyond which one cannot probe the curvature.
Is that reasonable? Are there any holes in my intuition? I'm mostly curious about that scenario because I never hear physicists talk about it (and I assume there's a reason:).
Bonus question: if this scenario is plausible, would that imply that Hawking radiation should cause a black hole to 'turn into' a neutron star at some point during the evaporation?
 A: The key flaw in your intuition is that once you are inside the event horizon, there is no amount of pressure or force that can keep you from moving further inwards. (It can be shown that the amount of force required to keep something at a constant radius goes to infinity as you approach the event horizon, and closer to the black hole than that it’s simply impossible to stay at a constant radius.) So you can’t say that the degeneracy pressure can maintain the surface of the star at a constant radius inside the event horizon. No amount of degeneracy pressure, no matter how great, can keep the particles in the star from collapsing down to a point.
A: No. Neutron stars (and quark stars, if they exist) are not black holes. There are many observational and theoretical differences. Some examples:

*

*Light is emitted from neutron stars (see eg: https://en.wikipedia.org/wiki/Pulsar), but cannot escape a black hole.

*There is a maximum spin of a neutron star before the star breaks up, which is much less than the maximum spin a black hole can achieve.

*Neutron stars are found to have masses in the rough range $1-3$ solar masses, while black holes can in principle have any mass.

The radius of a neutron star is larger the Schwarzschild radius. The ratio between the neutron star Schwarzschild radius and two times the radius of the neutron star is called the neutron star compactness.  If the neutron star was a black hole, the compactness would be 0.5, and larger neutron stars have smaller compactnesses. The compactness of a 1.4 mass neutron star with no spin (the "canonical" neutron star) is not known exactly because it depends on the equation of state of dense nuclear matter, which we don't know, but in most realistic models is predicted to be in the range 0.1-0.2; certainly much less than 0.5.
