This question involves the concept of "virtual particle" which was discussed a few days ago here. In a nutshell, a particle is virtual when it is a connecting line in a Feynman diagram between two vertices. It has all the quantum numbers of its name ( photon, electron, etc.) but not the mass, which is the measure of the four vector describing it. In that sense energy is not conserved with a virtual particle.
A virtual particle may continue out of the vertex and become "real", but a search for corresponding Feynman diagrams for Hawking radiation does not give any clear ones.
I found this heuristic one:
These loops are handwaved as coming "from the strong gravitational field in the hole .
This is the Feynman diagram of how pair creation is in the lab
It is necessary to have a second interaction because the created pair has invariant mass while the photon has zero mass.
One can adapt the left diagram to the heuristic loops of the Hawking plot above.
As the unification of gravity with the other three forces is not done yet, the diagrams are a guess that is missing an incoming "photon, Z0, gluon .." and an outgoing "photon, Z0, gluon..." ( do not now if gravitons can make a pair directly, were they unified with the rest) as the real particle in the feynman graphs. We do not know whether this will really work, and thus it is heuristic. I did find a publication that states the problem of feynman graphs and gravity for Hawking radiation.Anyway lets call the particle a generic "graviton".
If we assume that the graphs work ,then a "graviton" creates an e+e- pair which recombines within the horizon to the same "graviton". For certain conditions at the horizon one of the pair interacts with another "graviton" which supplies the second vertex, and falls into the hole while the other continues free. In this sense virtual is the e+ of the picture, whose second vertex is a "graviton" in the hole while the electron goes on as real picking up the energy from the potential of the hole. Thus black holes evaporate eventually in this scenario. No need for negative energies.
Best Answer
To start with, a photon is an antiparticle to a photon. There is no conservation of photons and certainly no negative energy photons.
Any body in space with a temperature larger than the "bath" it is in, i.e. the temperature of cosmic background radiation, will be radiating a black body radiation. The theoretical treatment is a combination of thermodynamics with field theoretical assumptions at the micrpscopic level, and that is the exposition in the wiki article..
The creation and annihilation of particles at the horizon is invoked to explain how particles can escape the gravitational attraction of the black hole at the microscopic level. As photons are antiparticles of photons and can be generated by in falling electrons for example, the only balances needed are directional and energy balances, so that the photon (or another particle) will not be trapped and fall back in.
In the case of the black body radiation of an ordinary body the temperature is such that the escaping radiation is a low energy photon, created by some transitions within the body and escaping from the surface. The energy balance is with the internal energy of the body, which cools incrementally. In the case of the black hole the energy balance is with its gravitational energy .
Here is a Feynman type diagram for the generation of particle/antiparticle by the Hawking radiation
To calculate the probabilities quantum mechanically would take exact diagrams. These pairs are virtual and they can also have photon vertices , which will have a probability of escaping the horizon and form a real photon spectrum from the black hole. For example: