There are basically two types of charged black holes, Reissner–Nordström, and Kerr–Newman.
Next, Hawking radiation itself is mostly electromagnetic. The numerous popular articles which say that Hawking radiation is particle/anti-particle pairs in which one of each pair gets swallowed by the black hole are misleading. Hawking radiation is almost entirely electromagnetic radiation---that is, photons. You can still use the language of particle/anti-particle, but it is a bit misleading.
This answer claims that Hawking radiation is just photons. Photons are EM neutral.
Now basically there are different interpretations of Hawking radiation, but what they mainly say is that the evaporation itself does not come from the black hole itself (nothing can escape the black hole), but from near outside the horizon.
Hawking radiation comes from the horizon, not from the black hole itself If you are worrying if your hypothetical black hole completely evaporates, i.e. no more horizon, it means that the electrons will just disperse due to the electrostatic repulsion, (once the body stops being a black hole).
Now these answers claim that BHs evaporate in a way that their leftovers are basically charged particles, that is, the charge that the BH did not evaporate stays during evaporation, until the BH's mass is not enough to create a event horizon anymore, and the BH ceases to exist anymore, and the leftover electrons just disperse.
Based on these it might be possible that the BH evaporates just photons, EM neutral particles, thus no EM charge is really evaporated from the BH, and the BH keeps its entire charge during the whole process of evaporation. At that point, as the BH's mass reduces so that it cannot create a event horizon anymore, the leftover electrons just disperse.
- Does a BH evaporate its EM charge?