As in the regular quaternions, we have the formula $(a_0+a_1i+a_2j+a_3k)(a_0-a_1i-a_2j-a_3k)=a_0^2+a_1^2+a_2^2+a_3^2$. So since the quaternions are associative, as long as this is nonzero then $a_0+a_1i+a_2j+a_3k$ cannot be a zero-divisor, as $\mathbb Z_2$ has no zero divisors.
It is elementary to check that four squares cannot add to $0$. Without loss of generality at least one is odd. Every square is equal to $0,1$ or $4$ mod $8$, and there is no way to add a $1$ and three $0$s,$1$s, or $4$s to be $0$ mod $8$.
No zero divisors means semicommutative and reversible.
In other words, since the Hamiltonian quaternions are ramified at $2$, $\mathbb H \otimes \mathbb Q_2$ is not isomorphic to the matrix algebra over $\mathbb Q_2$, so by Artin-Wedderburn it's a division algebra.