The answer is negative. Let me give a counter-example by modifying the example of singular complex algebraic space that is not a scheme given by Knutson in [algebraic spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is an elliptic curve $C=X_0$. The points $P_1,\dots P_9$ induce sections, let us choose the first one as the origin of the group law on $X_0$.

 Blow up $X$ in a tenth point $Q\in X_0$ chosen generic, so that $Q$ is not in the subgroup generated by the images of the nine sections. At this point, the strict transform of $X_0$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is a counter-example.

First, the fibers (and their infinitesimal neighbourhoods) are schemes by [Algebraic spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve in $Y$ intersecting the tenth exceptional divisor, but not meeting $y$. Its image in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, it meets $C$ only at $P_1,\dots, P_9$. This contradicts the fact that its stric transform in $Y$ should intersect the tenth exceptional divisor.