The answer is no. To help keep our notation straight, assume that there is a finite group $H$ and normal subgroups $G_1$ and $G_2$ of $H$ such that $G_1 \cong G_2$ and such that we have short exact sequences $$1 \longrightarrow G_1 \longrightarrow H \stackrel{f}{\longrightarrow} A_4 \longrightarrow 1$$ and $$1 \longrightarrow G_2 \longrightarrow H \stackrel{g}{\longrightarrow} D_6 \longrightarrow 1.$$ Choose this $H$ such that its cardinality is as small as possible.
Define $$\overline{G}_1 = G_1 / G_1 \cap G_2 \quad \text{and} \quad \overline{G}_2 = G_2 / G_1 \cap G_2.$$ The homomorphism $f$ induces an isomorphism between $\overline{G}_2$ and a nontrivial normal subgroup of $A_4$, and the homomorphism $g$ induces an isomorphism between $\overline{G}_1$ and a nontrivial normal subgroup of $D_6$. What is more, since $G_1$ is isomorphic to $G_2$ the groups $\overline{G}_1$ and $\overline{G}_2$ have the same cardinality. Examining the nontrivial normal subgroups of $A_4$ and $D_6$, we see that the only possibility is that these subgroups are actually the entire groups, i.e. that $$\overline{G}_1 \cong D_6 \quad \text{and} \quad \overline{G}_2 \cong A_4.$$ This implies that we have short exact sequences $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_1 \longrightarrow D_6 \longrightarrow 1$$ and $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_2 \longrightarrow A_4 \longrightarrow 1.$$ We conclude that $G = G_1 = G_2$ is a group that of cardinality strictly smaller than $H$ with fitting into the desired exact sequences, contradicting the minimality of the cardinality of $H$.
(the original post only contained a bunch of observations about the problem, but in the comments Ian Agol pointed out that one could use them as above to give a negative answer)