If yes, then how does this accord with relativity: the laws of physics are the same in all reference frames? We can move from a reference frame in which the photon has near zero energy density, to a reference frame where it has near infinite energy density.
If no, why are there no gravitational effects for a photon with an arbitrarily high energy density? For some reference frames, a photon can have more energy that the mass of the sun. How are these reference frames consistent with each other?
The answer to this question is through logical positivism--- the principle that the questions be formulated in a way that they give an experimentally realizable situation and ask "what happens"?
In this case, it depends on how you scatter things off the photon. If you scatter two of these things head on, they will combine to make a mini-black hole, and you need a full quantum gravity theory to describe the decay. But if you chase the photon and study it in a frame where its energy is very sub-Planckian, you will discover it is a photon. This is also the answer for a very glancing collision of two of these photons, moving nearly parallel.
The quantity which determines the regime which is appropriate is the center of mass energy of the collision.
Further, since all objects will produce black holes which are difficult to tell apart from one another, any highly boosted trans-Planckian point-particle undergoing a collision whose center-of-mass Schwartschild radius is bigger than the particle radius will look indistinguishable statistically from any other particle. This principle is the universality of the gravitational interaction at high energies, and it is an important focus of string theory research. Susskind investigated highly boosted strings on the path to black hole complementarity, concluding that a highly boosted string falling into a black hole will lengthen and wind around the horizon, and is indistinguishable from a surface excitation of the black hole.
