Here is an argument for $\operatorname{char} K=0$ (and probably also for $\operatorname{char} K>2$). Let $\pi\colon Q_8\to\operatorname{Sp}_{2n}(K)$ be an injective homomorphism, i.e., a vector space $V$ of dimension $2n$ with a faithful action of $Q_8$ and an invariant symplectic form.
The irreducible representations of $Q_8$ are $\{1,\chi_1,\chi_2,\chi_3,\rho\}$ where $\chi_i$ are order $2$ characters, and $\rho$ is the $2$-dimensional representation. Thus by complete reducibility $V$ decomposes as a direct sum of the five representations. However, the requirement that $Q/Z(G)$ is abelian says $-1\in Q$ acts as $-1$ on $V$, so in fact, only $\rho$ can appear in the decomposition of $Q_8$ (indeed, $-1$ acts trivially on all other representations). In other words, $V\simeq\rho^{\oplus n}$ as $Q_8$-representations.
Thus it remains to show that (up to conjugation) there is a unique symplectic form on $V$ compatible with the $Q_8$-action. Let $\omega\colon V\otimes V\to K$ be a symplectic form. Since $\rho$ also carries a $Q_8$-invariant symplectic form, we conclude there is a non-degenerate symmetric bilinear form on $\hom_{Q_8}(\rho,V)\simeq K^n$. Since we are working over an algebraically closed field, all non-degenerate symmetric bilinear forms must be conjugate to each other. Since there is an isomorphism $V\simeq \rho\otimes\hom_{Q_8}(\rho,V)$ of vector spaces carrying bilinear forms, we conclude that $V$ must have a unique non-degenerate symplectic form up to conjugation.
P.S. To answer LSpice's question: Since $Q_8$ is non-abelian, for $Q_8/Z(G)$ to be abelian, we need $Z(G)\cap Q_8$ to be non-trivial. Since $Z(G)=\{\pm I_{2n}\}$, this means $Z(G)\subset Q_8$. But since $Z(G)$ is central, it must be $\{\pm1\}\subset Q_8$.
