Generally speaking solids absorb light by converting the EM radiation to lattice vibrations (i.e. heat). The incident light causes electrons in the solid to oscillate, but if there is no way for electrons to dissipate the energy then electrons will simply reradiate the light and the light is reflected.
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.
But as dmckee says in his comment, the microscopic physics is reversible. If it's hard for oscillating electrons to transfer energy to bulk lattice vibrations then it's equally hard for those lattice vibrations to transfer energy back to the electrons and hence back out as light. So a shiny metal will be equally bad at absorbing and emitting light.
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.
In practice black body radiation is a mish mash of all sorts of different mechanisms, and the two cases I've mentioned are just examples. However in all cases when you look in detail at how energy is being transferred you'll find it's a reversible process and the energy flows equally fast in both directions.
You can't argue with a black object and a white object alone, as I think you partially understand in trying to build your thought experiment. You need a little bit more to define things properly. See whether the following helps.
Imagine a black object at a temperature $T_0$ and a white object also at $T_0$ inside a perfectly isolating box full of blackbody radiation at some higher temperature $T_1>T_0$ (i.e. without the black and white objects, this radiation is in thermodynamic equilibrium).
To understand exactly what would happen, you would have to describe the "colour" of your objects with emissivity curves that show emissivity as a detailed function of frequency. So your "black" and "white" would need to be defined in much more detail. You would also have to define the surface areas of the two objects and what they are made of (i.e. define their heat capacities). But all of this only effects the dynamics of how the system reaches its final state, i.e. these details only influence how the system evolves. What it evolves to is the same no matter what the details: the box would end up with everything at the same temperature such that the total system energy is, naturally, what it was at the beginning of the thought experiment. "Blacker" as opposed to "Whiter in this context roughly means "able to interact, per unit surface area, with radiation more swiftly": the blacker object's temperature will converge to that of the radiation more swiftly than does that of the whiter object, but asymptotically the white object "catches up". Blacker objects absorb more of their incident radiation its true, but they also emit more powerfully than a whiter object at the same temperature. The one concept emissivity describes the transfer in both directions. Think of emissivity as being a fractional factor applied to the Stefan-Boltzmann constant for the surface as well as being the fraction of incident light absorbed by the surface relative to a perfect blackbody radiator.
This description is altogether analogous to that of the situation where $T_0<T_1$. Begin with $T_0=T_1$, and you've got thermodynamic equilibrium from the beginning. Nothing happens, of course.
Maybe the following will help thinking about what is a really quite a complex question: it would be a fantastic last question for an undergrad thermodynamics exam BTW: You can abstract detail away by saying lets define object $A$ to be blacker than object $B$ if, when both objects are made of the same material, are the same size and shape, the temperature of $A$ converges to the final thermodynamic equilibrium temperature more swiftly than that of $B$ when they are both compared in the box-radiation-object thought experiment above.
Thinking about this now, I am not sure whether the above definition would hold for every beginning temperature of the radiation. Maybe there are pairs of surfaces whose relative blackness is different at different beginning temperatures such that $A$ is blacker than $B$ with some beginning temperature whilst the order swaps at a different beginning temperature. I think it is unlikely, but that is probably a different question altogether.
By the way, which pub do you drink in? I might come along.
Afterword on a Heater's Colour:
You ask by implication what is the best colour to paint a heater. This is not a simple question and involves the dynamics of the heater system. It's really an engineering question. I suspect in general it is better for them to be blacker rather than whiter. Here's a glimpse of the kind of factors bearing on the situation.
If you can say a heater has a constant nett input of $P$ watts, then at steady state that's going to be its output to the room, altogether regardless of its colour. There may be a materials engineering implication here: if you paint the heater whiter, and if its dominant heat transfer to the room is by radiation (rather than by convection or conduction), then it has to raise itself to a higher temperature than it would were it blacker so as to radiate $P$ watts into the room. So its materials might not be as longlasting, and it might be more of a fire hazard than it would be were it blacker.
If the heater is the hot water kind, and again if radiative transfer is significant, then the heating system has to run hotter to output power at a given level if the heater is whiter. At a given flow rate and given temperature of heating water, the heat output of heater is lower if it is whiter. You're trying to design the heater to be an "anti-insulator": you want the heat to leak out of the flow circuit in at the heater, not through the lagging on the hot water pipes outside the building channelling the water from the boiler to the heaters. If the hot water pipes leak heat in the same room, then that's no problem.
Recall the quartic dependence of the Stefan Boltzmann law. At room temperatures with a low temperature heater (the hot water kind) $\sigma\,T^4$ is likely to be pretty small compared with other heat transfer mechanisms, in contrast to my idealised scenarios above. So the heater's colour is likely to be pretty irrelevant.
Best Answer
The microscopic laws of physics all (excepting the weak interaction) have the property of being invariant on time-reversal. In a classical context that means that if I show you a very short film clip of pool balls colliding, you'll have a hard time knowing if I have shown it to you forward or in reverse. In quantum mechanics the meaning of "invariant" is a little different, but it implies that cross-section for the forward and reversed process are identical.
Now consider the emission and absorption of light. All possible modes are quantum mechanical interactions of the electro-magetic variety, which means that they are time-reversal invariant. So, any mechanism that emits photons of a particular wavelength efficiently from some high-energy also absorbs photons of the same wavelength efficiently from the lower-energy state. And those low energy states will be available because the thing is emitting all the time.
And the reversed argument also applies. Wavelengths that aren't emitted efficiently have no efficient mechanism for being absorbed either.
This is basically why a single quantity---the (wavelength dependent) emissivity serves to parameterize both emission and absorption.
This also points out one of the big puzzles of physics: how do microscopically reversible laws result in macroscopic irreversibility ala the 2nd Law of Thermodynamics. (And one answer comes from Statistical Mechanics.)