The analysis of one-particle states spanning a representation of the Lorentz group yields up to a phase convention

$$ \mathcal{T} b(E_p,\vec{p},s) \mathcal{T}^\dagger = (-1)^{1/2-s}b(E_p,-\vec{p},-s)$$

The change in the sign of spin is due to the fundamental relation $\mathcal{T} J \mathcal{T}^\dagger = -J$, so this conclusion holds for any kind of collective angular momentum. If $\sigma$ is some collective angular momentum, then the transformation above will apply, if not, you will have to look into the foundations of your model and derive the time-reversal from the transformation of the constituents.

The other question is, of course, what does $i$ stand for. If it is a position index, then it is untransformed, if some kind of fourier-transform/momentum index, then it has to be reverted (*without* touching the energy spectrum I assume would be somehow in matrix $U$).