Is an algebraic space over a DVR, whose special fibre and generic fibre are schemes, actually a scheme? Is an algebraic space over a DVR, whose special fibre (and all its infinitesimal neighborhood) and generic fibre are schemes, actually a scheme?
 A: The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface 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 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 blow-up 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 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 the counter-example we are looking for.
First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are one-dimensional [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.
A: 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 one-point 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.
