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There do in fact exist continuous martingales with stationary increments which are not Brownian motions. …

As $dM_t=M_{t-}\,d\tilde N_t$ and $d[M,S]\_t=M_{t-}\,d[\tilde N,S]\_t$, $M$ and $[M,S]$ are local martingales so, by integration by parts, $MS$ is a local martingale. … By localization (replacing $M$ by $M^\sigma$ for a suitable stopping time $\sigma$ with $\mathbb{P}(\sigma\ge\tau) > 0$), we can assume that $M$ and $MS$ are both proper martingales. …

there exists positive constants $c < C$ such that $$ c\mathbb{E}\left[[M]_t^{1/2}\right]\le\mathbb{E}\left[\sup_{s\le t}\vert M_s\vert\right]\le C\mathbb{E}\left[[M]_t^{1/2}\right] $$ for all cadlag martingales …

This example appears in Roger's & Williams book Diffusions, Markov Processes and Martingales as an example of a local martingale which is not a proper martingale. …

Yes, in this case it is true that $p$ is a proper martingale! Note that your integrand $\exp\left(-\int_0^tr_u du\right)$ is an adapted, continuous, and decreasing process bounded by 1. So, the follow …