In addition to Moishe Kohan's geometric argument, there's also a bordism-theoretic proof.

$\newcommand{\BDel}{B\mathrm{SL}_2(\mathbb C)^\delta}$ Let $\Omega_*^{\mathrm{SO}}(-)$ denote oriented bordism as a generalized homology theory. Your question is equivalent to asking whether $\Omega_*^{\mathrm{SO}}(\BDel) = 0$.

We can compute this with the Atiyah-Hirzebruch spectral sequence, which has signature $$ E^2_{p,q} = H_p(\BDel, \Omega_q^{\mathrm{SO}}(\mathrm{pt})) \Longrightarrow \Omega^{\mathrm{SO}}_{p+q}(\BDel). $$ $\Omega_q^{\mathrm{SO}}(\mathrm{pt}) = 0$ for $q = 1,2,3$, so in the range $p+q < 4$, this spectral sequence collapses, implying $\Omega_2^{\mathrm{SO}}(\BDel) \cong H_2(\BDel; \mathbb Z)$. Now, as suggested by Danny Ruberman's comment, Milnor's ``On the homology of Lie groups made discrete'' points out that $H_2(\BDel, \mathbb Z)$ surjects onto an uncountable $\mathbb Q$-vector space, hence is in particular nontrivial, and therefore not every $\mathrm{SL}_2(\mathbb C)$-representation of a closed, oriented surface extends to a compact $3$-manifold.

Unfortunately, this approach is difficult to make explicit for a given representation of a surface group without a better understanding of the homology of $\mathrm{SL}_2(\mathbb C)$ as a discrete group.

In addition to Moishe Kohan's geometric argument, there's also a bordism-theoretic proof.

$\newcommand{\BDel}{B\mathrm{SL}_2(\mathbb C)^\delta}$ Let $\Omega_*^{\mathrm{SO}}(-)$ denote oriented bordism as a generalized homology theory. Your question is equivalent to asking whether $\Omega_2^{\mathrm{SO}}(\BDel) = 0$.

We can compute this with the Atiyah-Hirzebruch spectral sequence, which has signature $$ E^2_{p,q} = H_p(\BDel, \Omega_q^{\mathrm{SO}}(\mathrm{pt})) \Longrightarrow \Omega^{\mathrm{SO}}_{p+q}(\BDel). $$ $\Omega_q^{\mathrm{SO}}(\mathrm{pt}) = 0$ for $q = 1,2,3$, so in the range $p+q < 4$, this spectral sequence collapses, implying $\Omega_2^{\mathrm{SO}}(\BDel) \cong H_2(\BDel; \mathbb Z)$. Now, as suggested by Danny Ruberman's comment, Milnor's “On the homology of Lie groups made discrete” points out that $H_2(\BDel; \mathbb Z)$ surjects onto an uncountable $\mathbb Q$-vector space, hence is in particular nontrivial, and therefore not every $\mathrm{SL}_2(\mathbb C)$-representation of a closed, oriented surface extends to a compact $3$-manifold.

Unfortunately, this approach is difficult to make explicit for a given representation of a surface group without a better understanding of the homology of $\mathrm{SL}_2(\mathbb C)$ as a discrete group.