As for **Q2**, I don't think so. For example, consider $(BG \times (\mathbb{A}^1\setminus \{ 0 \}))\amalg BG\to \mathbb{A}^1$, everything over $\mathbb{C}$.

**Added** [Edit: this one doesn't work, see comments below]: for a flat, but perhaps more contrived, example, I think one can take $\mathscr{X}:=B\mu_{3,\mathbb{Q}} \amalg B(\mathbb{Z}/3)_{\mathbb{Q}}$ and $X:=\mathrm{Spec}(\mathbb{Q})\amalg\mathrm{Spec}(\mathbb{Q})$, where $\mathscr{X}\to X$ is the coarse moduli space map. Here $\mu_{3,\mathbb{Q}}$ is the nonconstant group scheme over $\mathbb{Q}$ defined as a covariant functor on $\mathbb{Q}$-algebras by $A\mapsto$ third roots of $1$ in $A$, and $(\mathbb{Z}/3)_{\mathbb{Q}}$ is the constant group scheme defined on $\mathbb{Q}$-algebras by $A\mapsto\mathbb{Z}/3$ (Edit: this is for connected $Spec(A)$, in general it's a *locally* constant group scheme, as by the way it has to be a locally constant sheaf of groups).
Now, I think the two basechanges along the two possible maps $\mathrm{Spec}(k)\to X$, $k$ algebraically closed, are both the gerbe $B\mu_{3,k}=B(\mathbb{Z}/3)_k$.