There are hundreds of people who can answer this question better than I do here, but here is at least a partial answer. I hope that I do not misrepresent the truth too egregiously. (And I welcome corrections from people who are more deeply informed than I am.)

The (isomorphism classes of) finite-dimensional central division algebras over a field $K$ are classified by a gadget called the Brauer Group of $K$, and it is a basic fact of Local Class Field Theory that the BG of $\mathbb Q_p$ is isomorphic to $\mathbb Q/\mathbb Z$. The element of the BG corresponding to a central division algebra $H$ is called the *invariant of* $H$. So the answer to your question is, yes, there are plenty of division algebras over $\mathbb Q_p$, even ones whose center is $\mathbb Q_p$ itself. (That’s what one means when one speaks of a “central division algebra” over $\mathbb Q_p$.)

Here's a simple description of a cda of invariant $1/n$: let $U$ be the unramified extension of $\mathbb Q_p$ of degree $n$. I’m going to make an $n$-dimensional vector space over $U$ into a noncommutative ring. Let the basis be the set $\{1,\pi,\pi^2,\cdots,\pi^{n-1}\}$, where $\pi$ simply behaves like an $n$-th root of $p$: that is, $\pi^n=p$. And the question is how $\pi$ commutes with scalars from $U$. Here it is, $\pi u=u^\sigma\pi$, where by $u^\sigma$ I mean the image of $u$ under the Frobenius automorphism $\sigma$ of $U$ over $\mathbb Q_p$. Remember that the Galois group of $U$ over $\mathbb Q_p$ is cyclic, and $\sigma$ is a generator, satisfying the property that for $z$ in the ring of local integers of $U$, we have the relation $z^\sigma\equiv z^p\bmod{(p)}$.

If you’re really interested in this, I recommend that you try this out in the smallest case, $p=n=2$, and calculate the reciprocal of, for instance, $1+\omega+\pi$, where $\omega$ is a primitive cube root of $1$ over $\mathbb Q_2$, and $\pi^2=2$.