Interesting and complicated question. The things to consider:
"Black body radiation" assumes perfect absorption / radiation at all wavelengths. The greenhouse effect comes about from having absorption in the IR: the hot (short wavelength) radiation from the sun can penetrate the atmosphere, but the cooler earth radiates at a lower temperature - longer wavelength. And that longer wavelength light is reflected by the atmosphere (water, carbon dioxide, methane, etc). Think of the story of Winnie the Pooh visiting Rabbit's burrow. He goes in through the hole, has a "little smackerel" of honey (read: the whole pot), and then is too fat to get out again - commemorated on a postage stamp:
That's your photon. It had no difficulty penetrating the atmosphere as a short wavelength photon - but as a long wavelength photon it gets stuck as it tries to leave earth...
If you worry about your spaceship getting too cold (how big is it?) you should probably consider lowering its reflectivity - this directly scales with the heat loss. Notice how space ships are often "shiny metal". This isn't just because paint is expensive to lift into orbit (it is), but also to keep the emitted power down - keep the people inside protected from too much heat loss when not in the sun, and too much heat gain when they are. If you want to simulate the "Venus" effect you would want to create your own greenhouse effect - add a film that is transparent in the visible and opaque in the near IR.
Either way, your black body model needs to take account of the reflectivity as a function of wavelength - and instead of using the simple Stefan-Boltzmann law (which deals with total power per unit area), use the wavelength formulation (Planck law):
$$S_\lambda=\frac{2\pi hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$$
But yes - the amount of heat that a large object loses through radiation is substantial, even when it is at room temperature. There's a handy calculation on wolframalpha.com - it shows that the heat loss for an emissivity of 0.1 is still over $40 W/m^2$ at 298 K. The best thing you can do to insulate yourself is not let the outer shell get so hot in the first place - if you used a double shell, with the outer being thermally insulated from the inner, then you can see how this will lower the power emitted at the outer shell comes to equilibrium at some temperature $T_o$.
Assuming that the outer shell reflects half its power back to the inner shell, and half to the universe (which is so close to absolute zero that we ignore the difference), you can write
$$\epsilon \sigma T_i^4 = 2 \epsilon \sigma T_o^4$$
since the outer shell loses heat from both surfaces; thus if the inner shell is at $298 K$ the outer shell temperature will be at 250 K, but the inner shield is now losing heat at
$$\epsilon \sigma (T_i^4-T_o^4) = \frac12 \epsilon \sigma T_i^2$$
In other words - you halved it. If you add additional skins, the heat loss will be further reduced.
I must admit that I did that last bit of analysis "by the seat of my pants". It makes sense, intuitively, that the heat loss is reduced by a radiation shield; I have never attempted to come up with a number before, nor do I remember seeing this analysis. There could be a blooper in here - in which case I would be happy to have someone point it out.
I did find an online book that seemed to follow a similar approach but had a cylindrical geometry and uses different reflectivity on the inner and outer faces, which complicated matter further. But they show that multiple layers of shielding can significantly reduce these heat loads - which was really what I tried to say.
