Why is Hawking radiation composed of photons? The existence of the event horizon disrupts the vacuum state of quantum fields, and hence it appears to an observer that a black hole is generating radiation in its surroundings. Why is it the electromagnetic field that is perturbed? Why is "thermal" radiation generated? Can a black hole also generate perturbations in other fields?
 A: There are already some pretty good answers in this question, but I think there's still some things worth saying.

Why is it the electromagnetic field that is perturbed? [...] Can a black hole also generate perturbations in other fields?

All fields are perturbed. While most calculations are typically performed for the simpler case of a scalar field (which is the simplest), these results generalize. As a consequence, something particularly interesting that is currently being explored is that since Hawking radiation applies to all fields, it might also serve a purpose as a window into physics beyond the Standard Model. For example, there is work considering the production of dark matter particles by means of the evaporation of primordial black holes (see, e.g., arXiv: 2107.00013 [hep-ph]).
If you're interested in an intuition in general lines of how the effect works, you can check this answer I wrote a while ago.

Why is "thermal" radiation generated?

This is a bit trickier to answer without a "you do the maths and it comes out" approach lol. I'm not sure I can give you an adequate response, so I'll just list some other results that are related and make us believe that there is a deeper reason (I don't know this reason, and I don't know if someone does).
In classical general relativity, one can prove a few rigorous theorems about the behavior of black holes known as the laws of black hole mechanics. They are

*

*Zeroth Law: the surface gravity $\kappa$ of a stationary black hole is constant throughout its horizon

*First Law: if you perturb a stationary black hole, its mass $M$, area $A$, spin $J$, angular velocity $\Omega$, charge $Q$, and electric potential at the horizon $\Phi$ will satisfy
$$\mathrm{d}E = \frac{\kappa}{8\pi}\mathrm{d}A + \Omega \mathrm{d}J + \Phi \mathrm{d}Q$$

*Second Law: the area of the horizon of a black hole can never decrease, $\mathrm{d}A \geq 0$ (this is also known as the area theorem)

*Third Law: the surface gravity of a black hole can't become zero in finite time

As far as I know, these were originally shown by Bardeen, Carter, and Hawking in 1973 (zeroth through second law), and by Israel in 1986 (third law). The area theorem actually came a bit before the Bardeen–Carter–Hawking paper, but I don't really recall the original reference. Furthermore, I omitted some assumptions on the statements (for example, the reason black holes can evaporate and shrink is because quantum fields violate the weak energy condition, which is assumed in the area theorem).
In any case, notice how there are other four laws in physics that closely resemble these four:

*

*Zeroth Law: if two bodies are in thermal equilibrium with a third one, then they are in thermal equilibrium between themselves and have the same temperature

*First Law: energy is conserved in thermodynamical systems,
$$\mathrm{d}E = T \mathrm{d}S - \delta W$$

*Second Law: the entropy of a closed thermodynamical system can never decrease, $\mathrm{d}S \geq 0$

*Third Law: the temperature of a thermodynamical system can't reach zero in a finite number of steps

There is then a remarkable resemblance between the laws of black hole mechanics and the laws of thermodynamics. In particular, one see that both sets treat energy in the same way, that the area of a black hole seems to be an analogue of entropy, and that the surface gravity seems to be an analogue for temperature. In the Bardeen–Carter–Hawking paper, they dismiss this resemblance as a mere coincidence, since the temperature of a black hole obviously needed to be zero: you would never be able to put a black hole in thermal equilibrium with a bath at any other temperature, since the black hole doesn't emit anything. Hawking's prediction then showed that, once quantum effects are taken into account, the black hole's temperature will be given precisely by $T = \frac{\kappa}{2\pi}$ (units with $G = c = \hbar = k_B = 1$). Why is that the case? I have no better answer than "You make the computations and it shows up", but there is reason to believe there could be some deep connection between quantum mechanics, gravity and thermodynamics hidden in black holes.
A: The short answer to your title question is that it is not composed only of photons. Hawking's original paper, in which he derived the effect of particle creation by black holes, can be consulted here. In doing so, you will find that, in fact, he initially derived his result assuming a massless scalar field propagating in the background of a black hole formed by gravitational collapse of some matter distribution. This means he is studying massless particles of spin zero in such a spacetime. In this discussion he finds that an observer at the asymptotic future will observe this scalar field in a thermal state.
So we already see that in the original derivation, not only was there no claim that the result holds only for photons, but the result was in fact first derived for particles without any spin!
Later on he comments on spinning fields. You can find the comment in page (13) which I reproduce here:

Similar results hold for the electromagnetic and linearised gravitational fields. The fields produced on ${\cal I}^-$ by positive frequency waves from ${\cal I}^+$ have the same asymptotic form as (2.18) but with an extra blue shift factor in the amplitude. This extra factor cancels out in the definition of the scalar product so that the asymptotic forms of the coefficients $\alpha$ and $\beta$ are the same as in the Eqs. (2.19) and (2.20). Thus one would expected the black hole also to radiate photons and gravitons thermally. For massless fermions such as neutrinos one again gets similar results except that the negative frequency components given by the coefficients $\beta$ now make a positive contribution to the probability flux into the collapsing body. This means that the term $|\beta|^2$ in (2.27) now has the opposite sign. From this it follows that the number of particles emitted in any outgoing wave packed mode is $(\exp(2\pi \omega\kappa^{-1})+1)^{-1}$ times the fraction of that wave packet that would have been absorbed by the black hole had it been incident from ${\cal I}^-$. This is again exactly what one would expect for thermal emission of particles obeying Fermi-Dirac statistics.

His approach is fairly common: one often studies the simple case of spin zero first in full detail, and then comments on how the result generalizes to non-zero arbitrary spin $j$ by making the appropriate changes. As you can see in his comment, one gets the same result.
So the answer to your question is simply that the Hawking radiation is not composed only of photons. Hawking's original derivation predicts that whatever massless fields are propagating in the black hole spacetime will get thermal excitations produced by the black hole.
A: The electromagnetic field permits zero-mass excitations, down to arbitrarily low temperatures. Black holes are cold.
As the temperature approaches $kT≈1\rm\,MeV$, the high-energy parts of the blackbody spectrum can generate electron-positron pairs. In the final moments before evaporation, as the temperature becomes arbitrarily hot, presumably all of the quantum fields get involved.
A: A black hole formed from a neutron star has to emit anti-neutrinos too. As the star has formed by changing protons and electrons into neutrons and neutrinos, anti neutrinos have to be radiated away to make up for the the neutrinos that were produced (together with a lot of photons). The particle quantum numbers of the universe need to be conserved. It cannot be that after the evaporation there are too many neutrinos.
