Why does matter/antimatter only produce gamma rays?
At tree level, a matter-antimatter annihilation reaction doesn't just produce gamma rays, nor can you exclude neutrinos in the final state.
Even the simplest such reaction can—given enough energy—produce a variety of particle-pairs. However, those pairs are subject to two subsequent processes:
If the particles are not stable, they will decay towards lighter and more stable particles eventually reaching electron-positrons, nucleons/anti-nucleons, and neutrino/antineutrinos.
The anti-particles in the final state (excepting some of the anti-neutrinos) that don't decay will eventually find some matching particle to annihilate with.
So if we wait a while, the final state of the annihilation will be photons (mostly in the gamma band, but possible some low energy ones as well) and neutrinos (both matter and anti-matter (or both right and left handed if neutrinos are Majorana in nature)).
The neutrinos are often ignored because for practical purposes they have no effect other than carrying energy and momentum away from the interaction.
In the field of PET (positron emission tomography) people tend NOT to call the product of the annihilation "gamma rays", but rather "annihiliation photons". While that may be a subtle distinction, the view is that a "gamma ray" is emitted by a nucleus, while a "photon" is a more general term used for a quantum of electromagnetic energy. But according to Wikipedia, the definition / use of the term seems to depend on the field:
Electromagnetic radiation from radioactive decay of atomic nuclei is referred to as "gamma rays" no matter its energy, so that there is no lower limit to gamma energy derived from radioactive decay. This radiation commonly has energy of a few hundred keV, and almost always less than 10 MeV. In astronomy, gamma rays are defined by their energy, and no production process needs to be specified. The energies of gamma rays from astronomical sources range to over 10 TeV, an energy far too large to result from radioactive decay.[1]
(My emphasis).
So depending on the field - a photon is called a gamma ray if it has energy greater than 100 keV, or it is produced by nuclear decay. I think that no annihilation process will produce quanta less than 100 keV - but as @dmckee pointed out in his answer, there are annihilations where other particles (eg neutrinos) may be created.
You may ask "why not less than 100 keV?". That has to do with the mass of the particles. In the case of electron/positron pairs, their combined mass is equivalent to 1022 keV (2 x 511 keV). And if all matter disappears, it all has to turn to energy. It is possible though unlikely that an annihilation produces more than two photons - see this answer and links therein. This mentions that the electron-positron pair will annihilate into three photons in about 1/370th of all annihilations. I cannot find a description of the energy distribution in that case - but I imagine that the probability of one of these photons having energy less than 100 keV would be very small - especially in the c.o.m. frame of the annihilation (of course when your two particles are traveling near the speed of light relative to the observer, crazy things might happen to the energies of the observed particles.) This was measured in "Gamma-ray energy spectrum from orthopositronium three-gamma decay", Chang, Tang and Li, Phys Lett B, Volume 157, Issues 5–6, 25 July 1985, Pages 357-360 to which I unfortunately don't have access without handing over $39.95...
The amount of energy to emit is enormous ($2mc^2$ plus kinetic energy), so the emitted energies per photon will be at least the rest mass of the electron, which is half GeV (not counting some extreme and unlikely redshifts due to the collision point moving with respect to the observer). And that's well in the gamma spectrum.