Ok, so here is my answer to my own question:
Of the three options I presented in the question, the answer is 1: the emission of radiation into space is actually an irreversible process. At first I couldn't see how this could be the case, because the transfer of energy from the hot body to the outgoing radiation field doesn't involve a change in temperature, so it seemed like it should be reversible. What I had failed to consider is that the body interacts not only with the outgoing radiation field, but also with the incoming one (i.e. the cosmic microwave background in the case of a star.)
A useful metaphor is a heat exchanger, in which a pipe enters carrying cold water (temperature $T_C$), which is placed in contact with a body at a higher temperature $T_H$ until it equilibriates. Another pipe carries the warm water out.
Although the outgoing fluid is at the same temperature as the solid body, it is clear that this is an irreversible process. The entropy is produced not in the transporting away of warm water at $T_H$ but in the heating of water from $T_C$ to $T_H$.
A star may be seen as an analogous kind of "cosmic heat exchanger". In this case we have to imagine a small volume of space containing a small amount of cosmic microwave background energy at $3\:\mathrm{K}$ coming into contact with the star and being heated up to $6000\:\mathrm{K}$, and then moving away again.
These volumes of space should be thought of as moving at the speed of light. As with the heat exchanger, the transporting of the $6000\:\mathrm{K}$ radiation away from the star is a reversible process, but the "heating up of space" from $3\:\mathrm{K}$ to $6000\:\mathrm{K}$ is irreversible and produces entropy.
In my question I asked what would happen if this radiation were absorbed by a colder body, since it seems as if the entropy can decrease. For example, if $U$ Joules of thermal radiation at $4000\:\mathrm{K}$ (with entropy $\frac{4}{3}\cdot\frac{U}{4000} = \frac{U}{3000}\:\mathrm{JK^{-1}}$) were absorbed by a body at $3500\:\mathrm{K}$ then the body's entropy would increase by only $\frac{U}{3500}\:\mathrm{JK^{-1}}$ and it seems like the total entropy must have decreased. But if the second body can absorb all the radiation then it must be a black body, and hence it must emit black body radiation of its own according to the Stefan-Boltzmann law at a rate $A\sigma T^4$ (with A its surface area and $\sigma$ the Stefan-Boltzmann constant). It turns out that when you take the entropy of this outgoing radiation into account, the total entropy production is always positive. (Unless the absorbing/emitting body is at the same temperature as the radiation field, in which case it is zero, as we should expect in the case of thermal equilibrium.) This is like a heat exchanger operating the other way, with warm water coming into contact with a cold body, and cold water running out. If you forgot to take into account the entropy of the cold water it might seem as if the total entropy was decreasing.
Finally I should say why my argument about the outgoing radiation not doing work against a piston doesn't work. Imagine the following thought experiment. First we reversibly fill a piston with thermal radiation as described in the question. Then we make a small hole in the cylinder and let that radiation escape into space. This escaping radiation is exactly the same as black body radiation. The entropy of the radiation in the cylinder cannot decrease as it goes through the hole, and hence its entropy flux must be at least $\frac{4}{3}\frac{u}{T}$.
The problem with the Boltzmann definition is, as you have neatly shown, that its usefulness depends on the assumption that your system is in equilibrium with its surroundings. Without first assuming equilibrium and subsequently setting the temperatures as equal, one cannot show that the Boltzmann entropy satisfies the First Law and hence meaningfully define it as the entropy. However the Gibbs entropy still does provide a meaningful definition because, for example, it is possible to relate it to the partition function via
$$S \equiv -k_B\sum_s P_s\ln P_s= k_B \left(\ln Z + \beta\frac{\partial \ln Z}{\partial \beta}\right)=\frac{\partial}{\partial T}(k_B T \ln Z)$$
and hence to use it to calculate other variables like the Helmholtz free energy.
Furthermore, you are assuming that $W$ is large enough to be approximated as a continuous quantity. $W$ is however an integer and the quantity $\mathrm d\ln W$ is not well defined in a system with a small number of microstates. Hence the justification you have provided here would break down. This problem never arises if you start from the definition of the Gibbs entropy and work from there.
Best Answer
Your equation is correct only if:$$\mathrm dQ = mc\,\mathrm dT$$ which is not generally true, indeed, common sense tells you that a change in temperature leads to conclusion that an object being heated up. But we do not encounter gases much in our life, which could be regarded as a general case. In reality your assumption is generally false, a good example would be a general gas process: $$\mathrm dQ =\mathrm dW +\mathrm dU$$ Which would take this form for an ideal gas: $$P\mathrm dV+\frac{3}{2}NR\,\mathrm dT = P\,\mathrm dV \,\,\,\text{ if } \,\,\,\mathrm dT=0$$ Now you can see that although temperature does not change, heat supply is still possible, at least mathematically. This process is called Isothermal heating, as might have guessed already. This could serve you as an example of non increasing temperature heating.
After that, you should see that $dQ = mc\mathrm dT$ is usually not the case. Because temperature is not the only extensive parameter for a gas, i.e. is not the only thing which determines the energy supply, i.e. is not the only thing which rises when heats comes in. Real equation for entropy of an ideal gas then would look like: $$\mathrm dS = \frac{1}{T}(P\,\mathrm dV + (3/2)NR\,\mathrm dT) = NR\frac{\mathrm dV}{V} + \frac{3}{2} NR \frac{\mathrm dT}{T}$$
Then $S$ would look like,
$$S = NR \ln{ \left( \left(\frac{V}{V_0} \right) \left( \frac{T}{T_0} \right)^{3/2} \right)}$$
for $N = \textrm{constant}$, namely your number of particles does not change, and $V_0$ and $T_0$ play the same role as $C$ in your answer. Bear in mind that this result is not to be used in your studies, since some modifications and generalizations are to be made first, but it is good enough for your understanding.