Note that sheaf pushforward and preheaf pushforward agree, so this a question about categories, not sites.

**Lemma.** If $X$ is a finite type $k$-scheme with $\dim X > 0$, then $j_*h_X$ is not representable by a finite type algebraic space over $\mathcal O_k$.

*Proof (sketch).* Suppose $Y$ is a finite type algebraic space over $\mathcal O_k$ representing $j_*h_X$. This means that
$$\operatorname{Hom}_{\mathcal O_k}(T,Y) = \operatorname{Hom}_k(T \times_{\mathcal O_k} k, X)$$
for all $\mathcal O_k$-schemes $T$. If $T$ is any scheme of positive characteristic with a map $T \to \operatorname{Spec} \mathcal O_k$, then $T \times_{\mathcal O_k} k = \varnothing$. Thus, $\operatorname{Hom}_{\mathcal O_k}(T,Y)$ has exactly one element. Taking $T = \operatorname{Spec} \overline{\mathcal O_k/\mathfrak p}$ for $\mathfrak p \subseteq \mathcal O_k$ prime, we conclude that the special fibre $Y_\mathfrak p$ is zero-dimensional, as it is a finite type algebraic space over $\mathcal O_k/\mathfrak p$ with only one $\overline{\mathcal O_k/\mathfrak p}$-point. (For example, you can use Tag 06LZ and Tag 0ADC to prove this, plus a bit of work.) This actually shows that $Y_\mathfrak p = \operatorname{Spec} \mathcal O_k/\mathfrak p$.

But a finite type algebraic space over $\mathcal O_k$ whose closed fibres are zero-dimensional has all fibres zero-dimensional; for this you use that finite type schemes over $\mathcal O_k$ are Jacobson (hence the points over finite fields are everywhere dense). But now the generic fibre doesn't have enough points: if $T = \operatorname{Spec} \bar k$, then $\operatorname{Hom}_{\mathcal O_k}(T, Y) = \operatorname{Hom}_k(T, X)$ has infinitely many elements, as $\dim X > 0$. $\square$

There may very well be a way to construct $Y$ as a non-finite-type object. For example, if $X = \mathbb A^1$, then $j_* h_X$ is the quasicoherent sheaf $\tilde{k}$ on $\operatorname{Spec} \mathcal O_k$. You might be able to represent this as an algebraic space by writing down a resolution $\mathcal O_k^J \to \mathcal O_k^I \to k \to 0$ of $k$ as $\mathcal O_k$-module, and constructing $Y$ as the quotient of $(\mathbb A^1)^I$ by the equivalence relation defined by the presentation above. It might even be possible to glue this construction to treat the case of arbitrary (finite type) $k$-schemes $X$. (I didn't work this out because it sounds painful, and I'm not sure if this would be of any use to you.)