Briefly, for visibly transparent materials like glass, you can see through then even while they are glowing red hot. Most glasses have plenty of absorption in the IR, so there is plenty of absorption and re-emission going on. But once a visible photon is emitted, there's only a very low probability that it will be absorbed.
Will the visible emission still have the characteristic blackbody shape corresponding to the actual temperature of the material? It certainly looks that way to me, but how to understand the physics behind blackbody radiation from a transparent bodies?
In this really nice answer @RobJeffries explains the difference between the ideas of Thermal Radiation and Blackbody Radiation in the context of a source which may be in thermal equilibrium, but who's radiation may not be. It's worth a moment or two to read it before continuing here.
Here are some images from Wikipedia to help frame the question. The first from Blackbody illustrates the familiar cavity in radiative equilibrium with a small hole to sample that radiation.
If you visit a Glassblowing factory, studio, or demonstration, you are likely to see something like this, which is roughly similar (except for the flames).
The glass is introduce through the opening to be heated in the furnace, via some combination of absorption of the infrared light and contact with the hot gasses.
When it is pulled back out, the glass is usually glowing red, orange, or even yellow, depending on temperature,
...so that it softens and is easier to shape.
Assume for the purposes of this question that the glass has reached uniform temperature, removed, and the radiation from the glass measured. I believe that most blowing-friendly glasses used in this context are absorbing for at least a large chunk of the infrared.
When I watch the real thing, the glass visually appears to me to be transparent even when it is hot (it's really a beautiful effect!)
Question: Will the visible part of the radiation still approximate a Blackbody spectrum, even though the visible light does not experience several absorptions and re-emissions?
note: When it's very hot, it's very bright and so it's difficult to verify that the glass is still transparent, and I'm not going to start irritating someone holding hot glass by shooting lasers at them!