Why is a black hole's Hawking radiation temperature inversely proportional to its mass/size? As a black hole shrinks in volume and mass, shouldn't its temperature get lower?  Shouldn't it evaporate more slowly?
Naively, (very naively), I think that with a smaller surface area (as per its event horizon), it would have fewer and fewer random quantum foam particles popping up just outside its surface....
In a sense, you're right.  Vacuum energy over a surface area is proportional to the surface area so larger area means more of the particle-antiparticle occurrences and that's the source of Hawking radiation, that is, if you don't mind the virtual particle explanation, which is the only explanation I sort of understand.  I look at the equations and my brain goes blank.
But surface area isn't the only factor.  The more important factor is how quickly the gravitational field drops off, which, for stellar mass black holes or larger, the drop off over particle-anti particle distance is tiny, given a radius of at least a few miles and a very small particle-anti particle distance or high probability of a small distance.  That small drop-off in gravitational field leaves little chance for the outermost of the two virtual particles to escape.
That's why the only black holes that are thought to be hot are theoretical micro black holes, which are tiny.  The smaller the theoretical black hole, the faster it's gravitation falls with distance, so, even small distances like those between a particle and an anti-particle, make it much easier for one of the particles to escape the gravity, effectively stealing energy from the black hole and radiating outwards.