The answer is no if one makes the additional assumption that $S \in |-K_V|$ is ample, i.e. that $V$ is a smooth Fano threefold, as shown by the following example.
If $V$ is a Fano 3-fold containing $S$ as an ample divisor, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective (with torsion free cokernel).
Now, it is well-known that for all $g \geq 2$ there exists a polarized $K3$ surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $L$, see for instance
Picard group of a K3 surface generated by a curve
Therefore the injectivity of $r$ implies $\rho(V)=1$, or equivalently $b_2(V)=1$.
On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).
Therefore, if $g$ is large enough (it sufficies to take $g \geq 13$) we necessarily have $b_2(V) \geq 2$, a contradiction.
Remark 1. A. Beauville proved in this paper that a general $K3$ surface with given Picard lattice $R$ and polarization class $L \in R$ is an anticanonical divisor in a smooth Fano threefold if and only if there exists an isomorphism of polarized lattices $(R, \, L) = (\textrm{Pic}(V), \, K^{−1}_V)$ for some smooth Fano threefold $V$.
Remark 2. As observed in the comments below by D. Litt, J. C. Ottem and Mark, it is actually possible that a $K3$ surface $S$ appears as a non-ample anticanonical divisor in a smooth 3-fold, so the question in its general form is still unanswered.