Why do dark objects radiate thermal electromagnetic energy faster than light objects?

Kirchhoff's law of thermal radiation says that:

For a body of any arbitrary material, emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature, the perfect black-body emissive power.

I can imagine why dark objects have higher absorption of electromagnetic radiation: The darker the object is the less radiation it reflects back. An ideal black body would absorb all the incident electromagnetic radiation.

Is there a similarly simple and intuitive explanation of why dark objects emit thermal electromagnetic radiation faster that light objects and why is the Kirchhoff's law valid? For me it is not intuitive at all and I was not able to find any simple explanation.

• Emission is time-reversed absorption. Almost all of microscopic physics exhibits time-reversal symmetry (there are a very small number of energetic particle-physics processes which do not, and thermodynamics is famously irreversible). – dmckee Nov 12 '13 at 18:40

In metals the transfer of energy from oscillations of the conduction electrons to lattice vibrations is slow, so the light is mostly reflected. By contrast in graphite the light is absorbed by exciting $\pi$ electrons, and the excited orbitals efficiently transfer energy to the bulk so the light is mostly absorbed.
Similarly, in graphite if coupling of $\pi$ orbitals to lattice vibrations is efficient then energy flows equally fast both ways, and graphite will be equally good at absorbing and emitting light.