To simplify the situation let's imagine one wall of the room has an emissivity of 1 and the other has an emissivity of 0.1.
The question you're asking is whether you'll be able to see that the wall with emissivity of 0.1 looks darker. If so this does seem to be a paradox.
The resolution is that the emissivity and reflectivity are related by $E + R = 1$. Assuming everything is in thermal equilibrium the room and everything in is bathed in light with a colour temperature of 5000K. When you look at the wall with emissivity of 1 you see only emitted light because the reflectivity is zero. When you look at the wall with emissivity of 0.1 you see emission at only 10% of the other wall, but the other 90% is made up of reflected light. So the answer is that both walls will look equally bright.
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
A (perfect) white body, by definition, reflects all incident radiation and thus would not absorb any radiation from the sun.
If only considering radiative heating, yes. The white body will not heat up at all (see also 5).
No, see 2.
They will produce different spectra. The white body reflects all the incident radiation, so it's spectrum will be the same as that of the incoming sunlight. The black body, however, absorbs all incoming sunlight and emits blackbody radiation.
In general, there are more ways for heat transfer to occur than through radiation. Two other ways are conduction and convection. These are the main mechanisms by which objects reach thermal equilibrium with their surroundings. If you were to place two objects outside in the shadow, they would both eventually reach thermal equilibrium with the air (mainly through conduction and convection).