Does a body have to compress to its Schwarzschild radius to become a black hole? When something is compressed to within its Schwarzschild radius, it will compress to a singularity within finite time regardless of what forces are involved. But is there some lesser point at which neutron degenerate matter won't be able to support itself? It takes a lot of mass to make a Schwarzschild radius larger than an atom, but is that actually necessary to make a black hole?
 A: The answer you your question is no. There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse.
There are limits imposed by causality and General Relativity on the structure of compact stars. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro & Teukolsky, pp.260-261, it is shown, approximately, that even if the equation of state hardens to the point where the speed of sound equals the speed of light, that $(GM/Rc^2)<0.405$. 
The Schwarzschild radius is $R_s=2GM/c^2$ and therefore $R > 1.23 R_s$ for stability. This limit is reached for a neutron star with $M \simeq 3.5 M_{\odot}$.
A more accurate treatment in Lattimer (2013) suggests that a maximally compact neutron star has $R\geq 1.41R_s$.
If the equation of state is softer, then collapse will occur at smaller masses, and higher densities but at a similar multiple of $R_s$.
Thus it is not necessary to compress matter within $R_s$ to form a black hole.
The picture below (from Demorest et al. 2010) shows the mass-radius relations for a wide variety of equations of state. The limits in the top-left of the diagram indicate the limits imposed by (most stringently) the speed of sound being the speed of light (labelled "causality" and which gives radii slightly larger than Shapiro & Teukolsky's approximate result) and then in the very top left, the border marked by "GR" coincides with the Schwarzschild radius. Neutron stars become unstable where their mass-radius curves peak, so stable neutron stars are always significantly larger than $R_s$ at all masses. 

A: If the supernova remnant is less than about 3 solar masses, a neutron star will be the result. Neutron degeneracy will hold up the star, just as electron degeneracy holds up a white dwarf. Gravity wants to crush the neutrons out of existence, but neutrons are fairly solid, and push back. However, if the mass of the supernova remnant is greater than about 3 solar masses, the greater gravity will crush even the neutrons out of of existence. If this were to happen, the star would collapse further. In fact, if neutrons are crushed out of existence, then we know of no other force that can resist the crush of gravity. The star will, in theory, collapse all the way down to a black hole
