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.