In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$.  Since this is 0 for all n, it is 0 for $\tau$ as well, implying that $g(\tau \wedge t)$ is independent of the future.  I think this should make it $\mathcal F_t$ measurable.