= Solution
In sector $m$, separate the stationary winding path from its endpoint-fixed fluctuations:
$$
\phi(\tau)=\phi_0+\frac{2\pi m}{\beta}\tau+\eta(\tau),\qquad\eta(0)=\eta(\beta)=0.
$$
The cross term integrates to $2(2\pi m/\beta)[\eta(\beta)-\eta(0)]=0$. Hence
$$
\frac{I}{2\hbar^2}\int_0^\beta\dot\phi^2d\tau
=\frac{I(2\pi m)^2}{2\hbar^2\beta}+\frac{I}{2\hbar^2}\int_0^\beta\dot\eta^2d\tau.
$$
The Gaussian fluctuation integral is independent of $m$ and $\phi_0$. Denoting it by $\mathcal Z_0$, integration over the starting angle gives
$$
\boxed{\mathcal Z=2\pi\mathcal Z_0\sum_{m\in\mathbb Z}e^{-I(2\pi m)^2/(2\hbar^2\beta)}.}
$$
The <rotor return-kernel fluctuation prefactor> $\mathcal Z_0$ is the coincident-point free-particle kernel on the lifted line, or the return amplitude density from zero-net-winding paths. It is not the $n=0$ angular-momentum contribution: winding number and momentum quantum number label different descriptions.
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