As Piotr Achinger observes (but I shall stick to your notation), $\mathcal{M}$ has a coarse moduli space, say $M_0$, and we have morphisms $\mathcal{M}\to M:=\pi_0(\mathcal{M})^\mathrm{sh}\to M_0$. If $M$ is an algebraic space, then $M\to M_0$ has a section, hence $\mathcal{M}\to M_0$ must be an epimorphism for the étale topology (I assume here that $^\mathrm{sh}$ means "associated *étale* sheaf", but see below).

Here is an example where this fails (but at the moment I cannot think of one which is flat and quasifinite over a scheme). Let $k$ be a field of characteristic $\neq2$, and let $\mathcal{M}=[\mathbb{A}^2_k/G]$ where $G=\{\pm1\}$ acts in the obvious way. Then $M_0$ is the usual quotient scheme, and $\mathcal{M}\to M_0$ is an epimorphism iff the quotient map $f:\mathbb{A}^2_k\to M_0$ is. This is clearly not the case: this map has no section locally for the étale topology.

In fact, let me prove the stronger result that $\mathbb{A}^2_k\to M_0$ is not an epimorphism of fppf (or even fpqc) sheaves. In other words, we still have a counterexample if $^\mathrm{sh}$ means "associated fppf sheaf". Put $X=\mathbb{A}^2_k$, let $U$ (resp. $V$) be the complement of the origin in $X$ (resp. $M_0$), and $j:V\to M_0$ the inclusion. Then $\mathcal{A}:=f_*(\mathcal{O}_X)$ is a finite $\mathcal{O}_{M_0}$-algebra, étale of rank 2 on $V$ but not locally free on $M_0$ since its rank at the origin is 3. Moreover we have $$j_*(\mathcal{A}_V)=\mathcal{A}\quad\text{ and }\quad j_*(\mathcal{O}_V)=\mathcal{O}_{M_0}$$ because both $X$ and $M_0$ are $\mathrm{S}_2$. Note that these properties are *preserved by flat base change*.

Now assume $p:Z\to M_0$ is faithfully flat and lifts to $p':Z\to X$. Put $W:=p^{-1}(V)\subset Z$, and let $w:W\to Z$ be the inclusion. The étale double cover of $W$ induced by $U\to V$ has a section, so it is trivial and in particular $\mathcal{A'}:=p^*\mathcal{A}$ is free on $W$. Hence, using the above-mentioned base change property,

$$\mathcal{A'}\cong w_*\mathcal{A'}_W\cong w_*\mathcal{O}_W^2\cong \mathcal{O}_Z^2$$
which implies by descent that $\mathcal{A}$ is locally free over $M_0$, a contradiction.