Hawking radiation is emitted from the outside, if you start beyond the barrier it's usually not counted as an escape.

i thought that a black hole could eventually disappear through Hawking radiation.

Eventually being a really long time.

Yes, it could, and yes, it theoretically [in some models] contains/radiates information.

But the "radiation" happens right at the event horizon, the particles that get emerge have been "created" on the outer side.

https://www.youtube.com/watch?v=XT_iZ7QEit8

This seems as good a place as any to ask: when the particle-antiparticle pairs are made at the event horizon, one falls in, the other is emitted as radiation. Why don't the p & ap produced then pair up with and annihilate with other ap and p created in the vicinity such that it wouldn't really be detectable over ordinary vacuum activity. This could happen both inside and outside the horizon. It seems like Hawking radiation (if I've understood correctly) could be produced but would be almost impossible to detect as it would all annihilate immediately?

What have I missed?

Good question, but I have no idea, and ... it's ... complicated. Basically don't think about it in that way, because the math uses QM and frequencies.

https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles...

http://www.scholarpedia.org/article/Hawking_radiation#The_or...

The particle []or[] antiparticle (only one avoids falling in) produced usually does pair up and annihilate with another (anti)particle, but that happens above the event horizon, so the photon resulting from that annihilation sometimes is pointed away from black hole and escapes. It's only very small black holes where the pair-production particles themselves have sufficient height and speed to escape.