I'd like to show that
Let $R$ be a domain with the field of fraction $F$. Let $\mathcal{O}$ be an $R$-order in a $F$-algebra $B$. When $B$ is central simple, $Z(\mathcal{O})=R$.
I saw that $Z(\mathcal{O})=\mathbb{Z}$ for quaternion algebra $B$ over $\mathbb{Q}$, so I suppose this is true in other finite dimensions.
Since an order contains a $F$-basis of $B$, we have that $Z(\mathcal O) \subset Z(B)=F$.
I don't know whether it's now elementary to prove that $Z(\mathcal{O})=R$. Wedderburn's theorem can be used to show that $F$-algebra is quaternion if and only if it is central simple algebra of dimension $4$. I'm wondering if it might be helpful to prove this statement as well.