[edit:] Third law states 0K is impossible. That's the obvious and simplest answer. So everything bellow assumes the limit as you approach 0K.
I think in that case you reach a paradox because temperature is a quantity (typically) defined in thermodynamics, i.e. systems in equilibrium. A metastable glass is not in its ground state, therefore not in equilibrium, therefore not technically within the purview of thermo. This might seem like a cop out, but a similar question was asked in my qualifier. The answer, glass is not technically described my thermo and at 0K the whole thing breaks down [1]
Sure we still talk about entropy and temperature of glass, but it's stretching the definitions.
Another way to look at it, though, is that at 0K there is only one state available to the system (even though it is a glassy one). Therefore call the glass a new state of matter, and set S=0. If that feels weird because it's not the ground state, consider that glass' constituents, Si and O, are not in a ground state either, that'd be Fe. You don't have any problems dealing with metastable Si and O, do you? Either way, 0K makes no sense!
Also, it's weird (actually wrong) to even think about materials at 0K. In classical thermo your heat capacity is zero. In modern physics your atoms' "positions'" are fully determined, therefore their "momentum" is fully undetermined. So 0K is a state that makes zero sense.
[1] I forget the question. I think it was like this: the entropy of glass has a greater slope than the crystal, therefore, if you cool the glass low enough it will achieve a lower entropy than the crystal. How can a glass have lower S than its crystal state?
"A classical formulation by Nernst (actually a consequence of the Third Law) is:
It is impossible for any process, no matter how idealized, to reduce the entropy of a system to its absolute-zero value in a finite number of operations."