The textbook argument against photons canceling each other draws upon the conservation of energy. Does this mean that energy conservation is a "stronger" principle than superposition? Waves in other media than the EM field, e.g., sound or water, do cancel out---presumably by passing on their energy to some other degree of freedom (e.g., heat). Could this imply that EM waves don't have any alternative channel to pass on the destructed energy and thus can't cancel out?
You need to careful when you talk about photons as waves because one photon is certainly not an EM wave, which is a superposition of many many photons. Indeed if you are talking about very high energy photons they will produce an electrons positron pair. In other words you can convert the energy of photons into the "mass energy" (think about the famous equation $E = mc^2 \gamma$) of an electron positron pair. This process is called pair production and is entirely quantum mechanical. That's why you don't learn about it in your standard EM course.
The textbook argument against photons cancelling each draws upon conservation of energy.
Our theory of particle physics is called the standard model, and photons are point particles of zero mass , in the axiomatic table of the model,
and that is what the textbook is using. Yes conservation of energy is a very strict law. (In addition, photon photon interactions, and that is how elementary particles behave, are rare, for low energy photons, see here )
Does this mean that energy conservation is a "stronger" principle than superposition?
The classical electromagnetic wave when mathematically broken down into photons of energy hν (ν the frequency of the classical light) emerges in a complicated way from the quantum mechanical "addition" of the complex wave functions of each photon, the photon "wave" is a probability wave for each photon. (see this answer of mine, individual photons behave exactly the same way)
Waves in other media than the EM field, e.g., sound or water, do cancel out---presumably by passing on their energy to some other degree of freedom (e.g., heat).
Could this imply that EM waves don't have any alternative channel to pass on the destructed energy and thus can't cancel out?
Do not confuse classical EM waves with their component photons.A building may be made out of bricks, but a brick is not a building
We should be careful to distinguish interaction, correlation, annihilation and interference. Photons do not interfere. Any interference takes place at wave function level, so impacts the probability of finding a number of photons. Photons can annihilate but this requires two photons of a least 511 keV each in order to create an electron-positron pair. Photons can interact (scatter) via transient vacuum charge fluctuations. Finally photons can be correlated by Bose statistics.
All waves traveling through a medium don't cancel out. Sound waves, water waves, waves in a rope, etc. pass each other and travel further after they have passed. They don't (or almost not) exchange energy with the medium (like being converted to heat, though there is dampening). Two oppositely traveling waves may seem to vanish for a moment, but the medium contains the kinetic energy of the waves.
The same holds for classical em waves, although they don't travel in a medium. A circular em wave traveling outward from a center and a circular wave moving toward the same center will pass each other and continue their journey, without losing energy because they can't lose energy to the vacuum.
So they don't cancel out (they interact and travel along as if they didn't have encountered each other) because they don't have a channel to lose their energy.
It is a curious thing, because in the early days of quantum mechanics it was thought that wave interference effects (which is what I think you mean by cancelling) were only possible for a particle and itself (Dirac said exactly that, although I would have to put in a fair amount of work to find a reference).
This is indeed true for most particles in quantum mechanics. However, since the advent of quantum field theory, it has been recognised that the wave function for a photon does not describe the probability where where the particle may be found, but rather describes the probability for where it might be found that a photon has been annihilated. It actually does not have to be the same photon. Wave interference effects have actually been observed by astronomers for photons originating in different galaxies (sorry I don't have more in the way of details).
What I can say, is that your text book is out of date on this particular topic.