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
Perhaps the most direct example in particle physics is $J/\Psi$ (or any other meson not including up and down quarks) production.
The meson has a valence content of $c\bar{c}$, so it represent a pair of particles knocked on-shell from the nucleon sea.
The reaction is not exactly analogous because it requires a rather large input of kinetic energy (as does the electromagnetic pair production reaction ($\gamma \longrightarrow e^+ + e^-$) I showed in a comment now deleted.