When a star become black hole, does it attain a low entropy state? When a star goes supernova, its net entropy increases. 
If the left over neutron star then collapses into a black hole, does it attain a lower entropy state?
 A: There is the Bekenstein entropy bound that states a black hole is the maximum entropy that can be contained within a bounding surface of some area. Here the bounding surface is the event horizon
This is exactly what happens to a vacuum when observed from an inertial frame and from an accelerated frame. The transformation adjusts the vacuum to a vacuum with particles. For a temperature $T~=~ g/2\pi$ for $g$ the acceleration of the frame a change in the acceleration changes the temperature and thus the particle number. This is a form of $T~=~1/8\pi M$ for the temperature of a black hole.  The emission or absorption of a quanta of radiation adjusts the temperature with $M~\rightarrow~M~\pm~\delta M$.
The emission of a Hawking radiation particle is the transfer of entanglement phase from the black hole to particle states. The green hyperbolas are constant time surfaces and the other hyperbolic curves are constant distance. The red loop is an $O(\hbar)$ loop with the Hamiltonian $H$. The propagator for this is $e^{-iHt/\hbar}$ and we assign the euclidean time $t~\rightarrow~iħ/kT$, and around the loop the time sums to $t~\rightarrow~2\pi$. The partition function is then $e^{2πH/kT}$, and the Hamiltonian is defined by the radius of the loop, which is given by the hyperbola it intersects at distance $d~=~c^2/g$ above the horizon. The generation of the Hawking radiation, seen as the two red dots connected by a segment reduces the size of the horizon and the entanglement between the black holes in  two regions I and II. The observer in region I only witnesses one of the EPR pair and has no access to the other pair in region II. Hence this appears as a thermalization of the vacuum into vacuum plus radiation. The exact equation for the temperature is
$$
T~=~\frac{\hbar g}{2\pi kc}
$$

The equation for heat energy and entropy $dQ~=~SdT$ is used with $dQ~=~c^2dM$. The surface gravity of a black hole is 
$$
g^2~=~-\frac{1}{2}\nabla^aK^b\nabla_a\nabla K_b
$$
where the Killing vector $K_r~=~\sqrt{1~-~2m/r}\partial_r$ and $m~=~GM/c^2$ are used. This gives $g~=~c^4/4GM$ and so the temperature of a black hole is
$$
T~=~\frac{\hbar c^3}{8\pi GM}.
$$
We now compute the entropy and this is
$$
S~=~k\frac{\pi GM}{\hbar c}~=~k\frac{A}{4L_p},
$$
for $A$ the area of the event horizon and $L_p~=~\sqrt{G\hbar/c^3}$ the Planck length.
This illustrates that the entropy of a system contained within its Schwarzschild radius is the largest that can be contained in that area. One can think of this as meaning the complex information which made up the material contained in the black hole is now hidden and can be shuffled around within the black hole with no measureable change from our outside perspective.
A: No, it attains a much higher entropy state.
At least according to current understanding, black holes have the highest possible entropy for a given mass. 
If the entropy of the core material decreased upon gravitational collapse into a black hole, it would be thermodynamically favourable for it to convert back into a star. But of course this never happens.
