Is an algebraic space over a DVR, whose special fibre (and all its infinitesimal neighborhood) and generic fibre are schemes, actually a scheme?

The answer is negative. Let me give a counterexample by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.2122]. Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth 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 the elliptic curve $X_0=C$. Choose, as usual, an inflection point $O\in C$ as the origin of the group law on $C$. Let $\hat{X}$ be the blowup of $X$ in a tenth point $Q\in X_0=C$ (chosen generic, so that no multiple of $Q$ is in the subgroup of $C$ generated by $P_1,\dots, P_9$ : here, we use the uncountability of the base field). At this point, the strict transform of $X_0$ in $\hat{X}$ has negative selfintersection 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 the counterexample we are looking for. First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are onedimensional [Algebraic Spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not containing $y$. Its strict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, this plane curve would meet $C$ only at $P_1,\dots, P_9$. This contradicts the fact that $D$ intersects the tenth exceptional divisor. 


Olivier's example is perfect, but let me just point out that counterexamples are easier to construct if you allow nonseparated spaces. For example, start with two DVRs $R\hookrightarrow R'$ (with spectra $X'\to X$) such that $R'$ is finite free of rank $2$ over $R$, and the fraction field extension $K\hookrightarrow K'$ is separable. The constant group scheme $(\mathbb{Z}/2\mathbb{Z})_X$ acts naturally on $X'$. Let $G$ be the open subgroup scheme obtained by removing the nontrivial point over the closed point of $X$. Put $Y:=X'/G$. The morphism $Y\to X$ is an isomorphism on the generic points, but has the same closed fiber as $X'$. If $R\hookrightarrow R'$ is split unramified, then $Y$ is the familiar ``$X$ with the closed point doubled'' wich is a scheme. Otherwise, the closed fiber of $Y\to X$ is a onepoint scheme, which has no affine (or even separated) Zariski neighborhood in $Y$. In particular, $Y$ is not a scheme. You can check that $Y$ is locally separated (in Artin's sense: the diagonal map is an immersion) iff $R\hookrightarrow R'$ is unramified. 

