You can take the relative coarse map to get a factorization of $f$ into $\mathcal{X} \to X \to \mathcal{Y}$ where $g : X \to \mathcal{Y}$ is representable and $\pi : \mathcal{X} \to X$ is proper + quasi-finite with $\mathcal{O}_X \to \pi_*\mathcal{O}_{\mathcal{X}}$ an isomorphism. Then you can apply the representable case of ZMT to $g$ to obtain a factorization
$$
X \hookrightarrow \overline{X} \to \mathcal{Y}
$$
where $X \hookrightarrow \overline{X}$ is an open immersion and $\overline{g} : \overline{X} \to \mathcal{Y}$ is finite.

Putting this together, we get that any such $f$ factors into
$$
\mathcal{X} \xrightarrow{\rho} \overline{X} \xrightarrow{\overline{g}} \mathcal{Y}
$$
where

- $\overline{g}$ is finite and in particular representable,
- $\mathcal{X}$ is proper + quasi-finite over an open substack $i : X \subset \overline{X}$, and
- $\rho_*\mathcal{O}_\mathcal{X} = i_*\mathcal{O}_X$.

I think conditions $2 + 3$ can be replaced by something like $\mathcal{O}_\overline{X} \to \rho_*\mathcal{O}_\mathcal{X}$ is injective and integrally closed.

The existence of the relative coarse space is guaranteed under your assumptions by Theorem 3.1 here. Indeed the relative inertia stack is proper over $\mathcal{X}$ by the separated assumption and quasi-finite by the DM assumption.

I think by universality of the relative coarse map this is essentially the best you can do. In general the kernel of the map on inertia can jump so I don't think you can expect the first map to be a gerbe over an open substack, e.g., if $f$ itself the coarse space of a separated DM stack that is not a gerbe.