If you require your embeddings to be unital, then an embedding $\mathbb Z/p\mathbb Z \hookrightarrow D$ forces $D$ to have characteristic $p$, which means $\mathbb Z/q\mathbb Z$ cannot embed into $D$. Note that this doesn't assume $D$ is an integral domain.
If you do not require your embeddings to be unital, then it's impossible for essentially the same reason, but with a little bit more work; this is where you'll need to use the no zero divisors condition:
Suppose $\mathbb Z/p\mathbb Z \hookrightarrow D$, and say $1\in \mathbb Z/p\mathbb Z$ maps to $e \in D$. Then we get $p \cdot e = 0$ (where $p = 1+ \cdots + 1$, with $1 \in D$ being the identity). Since $D$ has no zero divisors, this implies $e=0$ or $p=0$. Since we assume $\mathbb Z/p\mathbb Z$ is embedded injectively, $e \neq 0$, so we have $p=0$, i.e., $D$ has characteristic $p$ (the characteristic cannot be zero, since $D$ is not the zero ring).