What is the temperature of dark matter?
No. Because it does not take part in the electromagnetic interaction, dark matter neither absorbs nor emits electromagnetic waves. This means that if dark matter particles do have a "temperature" then we do not have any direct means of measuring it.
These are actually two questions.
The temperature of the dark matter is an important value of many cosmological models. There are "hot" dark matter, "warm" dark matter and so on, models, proposing different energies of the proposed dark matter particles. This temperature (=kinetic energy of the particles) is an important factor for the dark matter behaviour even without the electromagnetic indication. Depending on their temperature, these particles, for example, may or may not have the escape velocity to get out of a galaxy, orbit the galaxy or just clump into the central black hole.
As for the infrared: whatever they are, the dark matter doesn't have much of a coupling to photons, so there is no electromagnetic emission of the dark matter that we are aware of. We can imagine, however, that if the dark matter decoupled early enough, its thermodynamic temperature is probably less than the microwave background. So even if there is a method of emission, it's probably not infrared either. It will be radio waves.
Dark matter might be in the form of thermal relics (e.g. WIMPs) or they may be produced non-thermally in phase transitions (e.g. axions). In the latter case, it does not make any sense to talk about a temperature at any stage.
For "thermal relics", the dark matter particles would thermally decouple from the rest of the universe as it cooled. Roughly speaking this occurs when the temperature in the universe falls below $\sim m_x c^2/k_B$ when other particles are not energetic enough to produce the dark matter particles. The density of dark matter particles would then drop to zero as they annihilated, but because the universe expands and becomes less dense, a state is attained where the dark matter particle interaction rate is too slow and they "freeze out". A detailed treatment shows that this happens at about $T_d\sim m_xc^2/20 k_B$.
The above would mean that the particles were non-relativistic when they decoupled, with an approximate Maxwell Boltzmann distribution at the decoupling temperature $\propto \exp(-p^2/2m_x k_BT_d)$.
From there, the distribution is fixed, but the particle momentum (with respect to the co-moving frame) decreases as the inverse of the scale factor, because it's de Broglie wavelength expands in the same way as that of light, and $p = hc/\lambda$. Thus the effective temperature of the dark matter goes as the inverse square of the scale factor.
This is quite different to the behaviour of particles that decouple while relativistic. For them, their momentum distribution is fixed e.g. $$ n(p) \propto (\exp [pc/k_BT_d] \pm 1)^{-1},$$ with the sign in the bracket depending on whether they are fermions or bosons. Here, you can see that the temperature of this distribution only decreases as the inverse of the scale factor as the universe expands.
The bottom line is that if dark matter was non-relativistic when it decoupled (e.g. WIMPS), then it is stone-cold now compared with, for example, the temperature of the photons (relativistic bosons) in the cosmic microwave background. The exact value would depend on the mass of the particle, but is not relevant, since they would behave like a pressure-free fluid under the influence of gravity.
Even neutrinos (with mass) as dark matter will be cold (now). That is because massless neutrinos would of course be relativistic and would have a similar temperature as the CMB, except that electron-positron annihilation after neutrino decoupling raises the photon temperature. However, neutrinos with mass decouple while still relativistic (for sensible neutrino masses), but become non-relativistic as the universe expands (when $p < m_\nu c$). As a result their temperature then declines more rapidly than the CMB.
The answer to your second question is that dark matter is called that because it does not have electromagnetic interactions. So it does not absorb or emit light of any wavelength.