In general $X$ is not a scheme, even if it is proper. But if $X\otimes_k L$ is a *quasiprojective* scheme, so is $X$.

First, assume $X\otimes_k L$ quasiprojective. By standard techniques, this still holds for some finite $L/k$. Then the projection $X\otimes_k L\to X$ identifies $X$ with the quotient (as fppf sheaf) of $X\otimes_k L$ by a finite locally free equivalence relation. The conclusion then follows from SGA3, V, Th. 4.1.

Now here is a counterexample where $X$ is proper, assuming $k$ has a separable quadratic extension $L$. We fix an algebraic closure $\overline{k}$ of $k$.

Hironaka has constructed a smooth proper $k$-variety $Y$ with a pair $\{y_1,y_2\}$ of $k$-points which is not contained in any affine open subscheme of $Y$ (not even after extension to $\overline{k}$). I claim that the Weil restriction $X:=R_{L/k} Y_L$ is not a scheme. Indeed, we have by definition $X(\overline{k})=Y(L\otimes_k\overline{k})=Y(\overline{k}\times\overline{k})=Y(\overline{k})\times Y(\overline{k})$. In particular we have a $\overline{k}$-point of $X$ corresponding to the pair $(y_1,y_2)$. If $x\in X$ is the corresponding closed point, one checks that $x$ has no affine neighborhood since this would give rise to an affine neighborhood of $\{y_1,y_2\}$.

On the other hand, $X$ is an algebraic space, proper over $k$ (in fact $X_\overline{k}$ is isomorphic to $Y_\overline{k}\times_\overline{k}Y_\overline{k}$).