Debarre's version of rigidity from Higher-dimensional algebraic geometry is
Lemma 1.15 Let $X, Y$ and $Y'$ be varieties and let $\pi: X \to Y$ and $\pi': X \to Y'$ be proper morphisms. Assume $\pi_* \mathcal O_X = \mathcal O_Y$. If $\pi'$ contracts one fiber $\pi^{-1}(y_0)$ of $\pi$, there is an open neighborhood $Y_0$ of $y_0$ in $Y$ and a factorization $$\pi'|_{\pi^{-1}(Y_0)}: \pi^{-1}(Y_0) \xrightarrow{\pi} Y_0 \to Y'.$$
Is there an example where this fails if $\pi'$ is not proper (while $\pi$ is still assumed to be proper)?
I think one still gets an open neighbourhood $y_0 \in Y_0 \subset Y$, such that all fibers $X_y = \pi^{-1}(y), y \in Y_0$ are contracted to points. However I'm not sure how to construct the factorization scheme-theoretically.
We will have to choose $X$ as a non-proper scheme (over $\mathbb Z$), since $Y'$ is reduced, and any morphism from a proper scheme to a reduced scheme is itself proper. Also I know that rigidity holds in the following situation (Mumford, Abelian varieties, p. 40):
Let $X$ be a proper variety, $Y$ and $Z$ any varieties, and $f: X \times Y \to Z$ a morphism such that for some $y_0 \in Y, f(X \times \{y_0\})$ is a single point $z_0 \in Z$. Then $f$ factors as $X \times Y \to Y \to Z$.
Here $f$ need not be proper, but one need the additional assumption of a product structure.