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More or less. In my layman's understanding: Black holes 'evaporate' slowly through Hawking radiation, losing mass as a function of their surface area (simplistically, particle/anti-particle pairs 'pop out of nothing' near the event horizon, one gets swallowed up the other escapes, this means a net loss of energy, which has to 'paid' by the black hole losing mass, think E=mc^2^).
Since a black hole behaves (geometrically) like any other sphere, the proportion of its area to its volume will grow as the black hole loses mass (i.e. it will have more and more relative area the smaller it gets), this process speeds up over time thus ending in what I guess you could call an explosion (more a whimper than a bang, to borrow a phrase).
Part 2 of your question: We don't know.
Wouldn't the hawking radiation need to be a higher rate than the black hole is absorbing matter?
Yes, the effect is extremely tiny and easily offset when a black hole is "feeding".
Which will eventually happen to all black holes because the last things remaining will be black holes, so there would be no matter to absorb.
Which begs the question, what happens to the estranged particle that escapes the black hole from hawking radiation.
They'll wander forever through an ever expanding space, meaning they probably won't ever come across a different particle.
Eventually everything will reach equilibrium, aka the state where nothing moves anymore because everything it could react with is too far away to cause any reaction.
Which is why it would work with a small black hole, but not with a large one