There exist hyperbolic 3-manifolds which cannot embed totally geodesically in complex hyperbolic manifolds, answering this question in the negative.
Recently it was shown that complex hyperbolic manifolds have integral discrete faithful representations; in particular, the traces of the matrices in $SU(n,1)$ have integral traces. See Theorem 1.3.1 of Baldi-Ullmo or Theorem 1.5(3) of Bader et. al..
This follows from a result of Esnault-Groechenig that cohomologically rigid representations of the fundamental group of smooth projective varieties must be integral. Compact hyperbolic $n$-manifolds are projective varieties, and the discrete faithful representation into $SU(n,1)$ is unique up to conjugation and cohomologically rigid by Mostow rigidity. Hence this representation must have integral traces (this corollary was pointed out to me by David Fisher, but see the above citations for more details).
However, there are hyperbolic 3-manifolds such that the discrete faithful representation of the fundamental group into $SO(3,1)$ has non-integral traces, implying that they cannot embed isometrically in a complex hyperbolic 3-manifold. See Theorem 1.8.