Yes: $\mathrm{Sp}(2n, \mathbb F_p)$ is generated by its root subgroups. Each root subgroup is cyclic, and generated by an element that lifts in an obvious way to $\mathrm{Sp}(2n, \mathbb Z)$.

More concretely, if I haven't messed up the calculations in your form (I'm used to a different one), then, writing $E_{i j}$ for the matrix that has a 1 in its $(i, j)$th entry and 0 elsewhere, we have that $\mathrm{Sp}(2n, \mathbb F_p)$ is generated by:

- $\{1 + t(E_{i j} - E_{(i + n)(j + n)}) : t \in \mathbb F_p\}$ (corresponding to the root $e_i - e_j$) for $1 \le i < j \le n$,
- $\{1 + t(E_{i(j + n)} + E_{j(i + n)}) : t \in \mathbb F_p\}$ (corresponding to the root $e_i + e_j$) for $1 \le i \ne j \le n$,
- $\{1 + t E_{i(i + n)} : t \in \mathbb F_p\}$ (corresponding to the root $2e_i$) for $1 \le i \le n$,
and
- the transposes of the various groups above (corresponding to the negative roots).