The answer is yes. By the $L^2$ bound on $\mu$ and Doob's inequality, $M_t$ is $L^2$ bounded, and a fortiori uniformly integrable. Hence it is closable - that is, $M_t = \mathbb E[M_1 | \mathcal M_t]$, where $\mathcal M_t$ is the natural filtration of $M$.
Now let $X$ be such that $\text{Law}(X) = \text{Law}(M)$ and $X_1, Y_1$ be an optimal coupling of $\mu$ and $\nu$. By definition we then have $\mathbb E[|X_1 - Y_1|^2] \leq \varepsilon^2$.
Define the martingale $Y$ by $Y_t := \mathbb E[Y_1| \mathcal X_t]$, where $\mathcal X_t$ is the natural filtration of $X$. Then $X - Y$ is an $\mathcal X_t$ martingale, and we have by Doob's inequality
$$\mathbb E[\sup_{0 \leq t \leq 1} |X_t - Y_t|^2] \leq 4 \mathbb E[|X_1 - Y_1|^2] \leq 4\varepsilon^2,$$
as desired.