There are several places for the energy to go. One of the simplest places its lost is to reflections. To get out of the the laser housing, the light has to leave the IR diode, passing into and then out of a lens, enters and then leave 2 more crystals, and then enters and leaves whatever collimating lens you're using. That's 8 interfaces, double the number of interfaces it has to pass through compared to using just a diode (4 times the number with an open-can), so that's double (or 4 times for an open-can) the losses of what you'd get with just a diode. This is because some light is always reflected when passing through an interface. So that is one source of losses, but that's not the major one.
Another answer to where energy goes is basically what you said, heat. Photons can't just be converted to another wavelength without any energy change, because different wavelength photons have different energies. In fact, a 532nm(green) photon has more energy than a 808nm(IR) photon. These energy conversions between photons and electron transitions in the crystals are not completely efficient, and energy is lost through these transitions and through electron and atom vibrations caused by the photons passing by. An atomic vibration, also known as a phonon, is exactly heat.
So basically, it's just inefficiencies, which either don't let the light through or turn their energy into heat in the crystals. Some higher-powered lasers do let IR through, which is why you need a filter to make sure there's no IR leaking through. The filter will simply absorb the photons and the energy will be dissipated into heat.