First note that if $A$ is the $2\times 2$ matrix algebra over a field, and $z\in A$ has trace zero, then by the Cayley Hamilton Theorem, $z^2=-det (z)$ is a scalar. Hence, if $x,y \in A$ then $(xy-yx)^2$ is a scalar matrix.
Suppose $D$ is a skew field, and is finite dimensional (of dimension $n$) over its centre $K$. We may assume that $D$ is not commutative. Then the centre cannot be a finite field and hence $D$ is Zariski dense in $D\otimes _K {\overline K}\quad $ (${\overline K}$ is the algebraic closure of $K$). Then if $dim (D)\geq 3$, then $(xy-yx)^2$ is not in $K$ for some $x,y\in D$.
Your condition says that for all $x,y\in Q$ we must have $(xy-yx)^2$ lies in the centre of $Q$. Hence the same is true for $D$. Hence $D$ is indeed quaternionic.