The thermal radiation associated with some object is typically described in terms of the "black-body" spectrum for a given temperature, given by the Planck formula. This formula is based on an idealization of an object that absorbs all frequencies of radiation equally, but it works fairly well provided that the object whose thermal spectrum you're interested in studying doesn't have any transitions with resonant frequencies in the range of interest. As the typical energy scale of atomic and molecular transitions is somewhere around an eV, while the characteristic energy scale for "room temperature" is in the neighborhood of 1/40 eV, this generally isn't all that bad an assumption-- if you look in the vicinity of the peak of the blackbody spectrum for an object at room temperature, you generally find that the spectrum looks very much like a black-body spectrum.
How does this arise from the interaction between light of whatever frequency and a gas of atoms or molecules having discrete internal states? The thing to remember is that internal states of atoms and molecules aren't the only degree of freedom available to the systems-- there's also the center-of-mass motion of the atoms themselves, or the collective motion of groups of atoms.
The central idea involved with thermal radiation is that if you take a gas of atoms and confine it to a region of space containing some radiation field with some characteristic temperature, the atoms and the radiation will eventually come to some equilibrium in which the kinetic energy distribution of the atoms and the frequency spectrum of the radiation will have the same characteristic temperature. (The internal state distribution of the atoms will also have the same temperature, but if you're talking about room-temperature systems, there's too little thermal energy to make much difference in the thermal state distribution, so we'll ignore that.) This will come about through interactions between the atoms and the light, and most of these interactions will be non-resonant in nature. In terms of microscopic quantum processes, you would think of these as being Raman scattering events, where some of the photon energy goes into changing the motional state of the atom-- if you have cold atoms and hot photons, you'll get more scattering events that increase the atom's kinetic energy than ones that decrease it, so the average atomic KE will increase, and the average photon energy will decrease. (Or, in more fully quantum terms, the population of atoms will be moved up to higher-energy quantum states within the box, while the population of higher-energy photon modes will decrease.)
For thermal radiation in the room temperature regime, of course, the transitions in question are so far off-resonance that a Raman scattering for any individual atom with any particular photon will be phenomenally unlikely. Atoms are plentiful, though, and photons are even cheaper, so the total number of interactions for the sample as a whole can be quite large, and can bring both the atomic gas and the thermal radiation bath to equilibrium in time.
I've never seen a full QFT treatment of the subject, but that doesn't mean much. The basic idea of the equilibration of atoms with thermal radiation comes from Einstein in 1917, and there was a really good Physics Today article (PDF) by Dan Kleppner a few years back, talking about just how much is in those papers.
The electromagnetic processes between atoms and molecules in all phases, solid, liquid, gas, depend on what is generically called "Van der Waals" fields and subsequent forces.
It is well known that the atoms/molecules are neutral, nevertheless there exist for all matter dipole and quadrupole and higher order fields which are mainly attractive and form the chemical bonds which is the way neutral atoms and molecules can bind into solids and liquids and interact as gases.
These bonds are quantum mechanical, that means that there exist solutions of the Schrodinger equation with energy levels from ground state to continuum, one can model them as repeated over all the mass of the solid , liquid and gas. The unfilled energy levels are close to each other in energy and the continuum of n=infinity ( the radial quantum number).
At the same time the atoms and solids have pure kinetic degrees of freedom: they can vibrate and rotate in solids, they can move in two dimensions in liquids and in all three dimensions in gases.
In gases simple scattering of the molecules transfers the kinetic energy of one molecule to the potential energy of another, i.e. raises an electron to a higher level. The electron goes back to its ground state releasing a specific photon, or a cascade of photons, depending on the energy. Remember that the higher levels with respect to n, the radial quantum number, are closely packed. These photons are the ones emitted as black body radiation, and they are a continuum because of the 10^23 molecules per mole and the almost continuous energy levels. The temperature is a function of the average kinetic energy in the gas, the higher the temperature the more energetic the kinetic scattering and the higher the average photon energy.
In a solid there are also vibrational and rotational kinematic degrees of freedom that are contributing to the average kinetic energy, i.e. temperature. The kinetic energy of the molecules becomes potential energy for an electron in the lattice which then decays to its ground state or through cascades. The logic is the same as for gases and the same holds for liquids that have some extra kinematic degree of freedom with respect to solids.
So it is the quantum mechanica behavior of matter at the micro level which is responsible for the black body radiation, and the infrared catastrophy problem of the classical extrapolations was solved. It is the energy levels that make a difference between infinity and well behaved property in the electromagnetic emmissions. Thus the average kinetic energy( proportional to T) diminishes by turning into electromagnetic emissions through the stepping of energy levels.
Does that mean in rarefied conditions where the mean-free-path is relatively large, the rate of IR emissions decreases (while the intensity is still only dependent on the temperature)?
When the mean free path is large, the temperature is lower, the average kinetic energy of the atoms is lower and thus the photons produced by the transformation of kinetic to exciting electrons to higher potential energy levels and consequent decay to ground energy levels are all lower and will keep getting cooler if the energy is not replenished. I do not know what you mean by the intensity.
This is the black body radiation spectrum.
As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.
If you mean gases in low pressure as at the top of the atmosphere etc, one has to study them separately according to the boundary conditions. There can be gases with very high temperatures as in the atmosphere of the sun.
Best Answer
Blackbody radiation is an idealized description of thermal radiation of a substance which is in thermal equilibrium with the photon field. Your description in the question, which equates thermal radiation and blackbody radiation is therefore not quite accurate. Indeed, blackbody radiation is quite simple compared to thermal radiation in general -- blackbody radiation doesn't depend at all on material properties, all it depends on is the temperature and fundamental constants of the electromagnetic field. Why is this? Philosophically, when two things are in thermal equilibrium, we can "set an equals sign" between many of their properties. Therefore, when an object is in thermal equilibrium with the field of photons, to understand the radiation it emits, we only need properties arising from the statistical mechanics and spectrum of photons in three dimensions.
Johnson noise also is independent of the properties of the material, and as it also has to do with electromagnetic fields, one might expect it to be related to blackbody radiation inside the conductor. This is indeed the case, but now we must concern ourselves with photons in one dimension, since the typical setting for Johnson noise is in a wire, rather than free space! This explains the difference in the formulas. Derivations of Johnson noise in this context can be found here and here (just from a google search of "Johnson noise" and "blackbody").
You can read a discussion of how to think about Johnson noise in terms of blackbody radiation in this classic 1946 paper by Robert Dicke titled "The Measurement of Thermal Radiation at Microwave Frequencies". The physical point made there is that an antenna receiving blackbody radiation at temperature $T$ and a resistor at temperature $T$ experiencing Johnson noise must have equal power. The difference in the forms of the power spectra is apparently due to the frequency dependence of the antenna's detection pattern.