Being a sheaf in the étale topology is insufficient. Begin with the locally ringed space $(\text{Spec}\mathbb{Z},\mathcal{O}_{\text{Spec}\mathbb{Z}})$. Let $$i:X\hookrightarrow \text{Spec}\mathbb{Z}$$
be the complement of the generic point $\langle 0 \rangle$. Give $X$ the sheaf of rings $\mathcal{O}_X := i^{-1}\mathcal{O}_{\text{Spec}\mathbb{Z}}$. Let $$i^\#:\mathcal{O}_{\text{Spec}\mathbb{Z}} \to i_*\mathcal{O}_X,$$

be the natural map defined by adjunction of $i^{-1}$ and $i_*$. Then $(i,i^\#)$ is a morphism of locally ringed spaces. Moreover, $(i,i^\#)$ has the following universal property: for every locally ringed space $(T,\mathcal{O}_T)$ and for every morphism of locally ringed spaces
$$ (f,f^\#):(T,\mathcal{O}_T) \to (\text{Spec}\mathbb{Z},\mathcal{O}_{\text{Spec}\mathbb{Z}}),$$
(of course there is a unique such morphism!), there exists a morphism of locally ringed spaces
$$ (g,g^\#):(T,\mathcal{O}_T) \to (X,\mathcal{O}_X),$$
with $(i,i^\#)\circ(g,g^\#) = (f,f^\#)$ if and only if $g(T)$ is contained in $i(X)$, in which case the morphism $(g,g^\#)$ is unique. To learn more about any of this, please read about the "max spectrum".

Now consider the Yoneda functor $h_X$ of $(X,\mathcal{O}_X)$, but restricted to the full category of schemes. This is simply the functor that associates to every scheme $(T,\mathcal{O}_T)$ the singleton set if every residue field has positive characteristic, and that associates the empty set if some residue field has characteristic $0$. This is an étale sheaf. It is representable by the empty set if you restrict to the category of $\mathbb{Q}$-schemes. However, it is not representable by a scheme or algebraic space, not even when restricted to the category of $\mathbb{Z}[1/n]$-schemes. To see this, let $p$ be any prime not dividing $n$, and consider the value of the functor restricted to the sequence of schemes $$(\text{Spec}\mathbb{Z}/p^r\mathbb{Z},\mathcal{O}_{\text{Spec}\mathbb{Z}/p^r\mathbb{Z}})_r$$
as well as the scheme $(\text{Spec}\widehat{\mathbb{Z}}_p,\mathcal{O})$. Since the functor fails to be *effectively* prorepresentable, it is not a scheme or algebraic space.