To the extent I know, before Bekenstein and Hawking, the general consensus among physicists was that the black holes have zero temperature and zero entropy. I can possibly understand why people would have thought that it has zero temperature because anything with a nonzero temperature would radiate and I believe, that seemed odd with the idea of a classical black hole. But why did they believe that the black holes have nonzero entropy? Does having zero temperature mean having zero entropy?
3 Answers
They didn't associate entropy with a black hole at all. They had no intuition about it at all, basically because they saw no internal structure to a black hole and evaluating entropy requires counting the number of micro-states. It was the intervention by Bekenstein that raised this question and in fact he was able to write out Hawkings entropy formula without the proportionality formula. This formula is correctly called the Bekenstein-Hawking fornula.
Moreover, Hawking first attempted to prove Bekenstein wrong in a semi-classical calculation but in the end, to his surprise, was forced to the conclusion that he was right. He even managed to derive the proportionality constant.
Of course then the question was raised as to what were the microstates that the formula was counting. In Loop Quantum Gravity, a contender for quanyum gravity, it turns out it's counting the number of quantum geometrical states and in String Theory, another contender for quantum gravity, it turns out it is counting BPS states around a configuration of D-branes - however, this calculation has only been done in higher dimensions, if memory serves it was 5d spacetime.
A classical blackhole does not contain matter. The Schwarzschild metric is a vacuum solution, meaning the blackhole is made from curved spacetime: A stable gravitational soliton completely described by one parameter, $M$.
So it's static, contains nothing, and is described entirely by one constant number (with dimensions of mass), for all of eternity. "Entropy" is not the first thing that comes to mind. It appears to be $\ln{(1)}=0$...like asking, what is the entropy of an alpha particle.
Of course, quantum considerations changed everything. It made them into dynamic objects with temperature, entropy, and time evolution.
The important questions is whether zero temperature equals zero entropy. And this is in fact the content of the third law of thermodynamics, which states that the entropy of a system at zero temperature is equal to a well-defined constant. From a microscopical point of view this constant is the log of the number of degenerate ground states.
For crystalline materials (and a host of other things) the ground state is unique and the residual entropy at zero temperature is in fact zero. But for some systems, mostly glasses, there is a number of different states of minimal energy and this gives rise to a positive residual entropy.
As for black holes, it was known previous to Bekenstein that a black hole should be assigned some entropy. That is because if the entropy of a black hole were zero (even in the context of classical relativity) then it would be possible to violate the second law of thermodynamics. All you need to do is to toss some random system with positive entropy in a black hole, once it traverses the event horizon you would lose all the entropy there contained. In fact it was this particular discussion that led spurred Bekenstein.
As a side note, Kerr black holes have a Hawking temperature that depends on both the mass and the angular momentum. For a extreme Kerr black hole (one in which $J=M^2$ in gerometric units) the Hawking temperature equals to zero, but the horizon area stays finite, so even for black holes it is not the case that zero temperature gives zero entropy