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 

http://mathoverflow.net/questions/124880/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][1] 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 remarked 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.


  [1]: http://math1.unice.fr/~beauvill/pubs/Fano.pdf