Yes, one can always do this. In fact, one can assume that $\det(U)=1$, as this follows by a homotopy argument, using the fact that $\pi_3(S^2)\simeq \mathbb{Z}$.

Here are the details: Assume that $\det(U)=1$ so that $V$ has a non-negative real determinant. Then one can write $V$ uniquely in the form
$$
V(x) = \begin{pmatrix} x_0+i\,x_1 & -x_2 + i\,x_3\\
x_2-i\,x_3 & \phantom{-}x_0-i\,x_1
\end{pmatrix}
$$
for some $x = (x_0,x_1,x_2,x_3)\in\mathbb{R}^4$.

Now consider the map $F$ from $\mathbb{R}^4$ into $\mathbb{R}^3$ (regarded as the traceless Hermitian $2$-by-$2$ matrices) defined by
$$
F(x) = \left[A +BV(x) + V(x)^*B^* + V(x)^*CV(x)\right]_0\,,
$$
where, for a Hermitian $2$-by-$2$ matrix, $H$, we write $H - \tfrac12\mathrm{tr}(H)\,I_2 = H_0$. For use below, define the norm on traceless Hermitian $2$-by-$2$-matrices $M$ by the rule $|M|^2 = \tfrac12 \mathrm{tr}(M^2)$.

Our task is to show that there exists an $x\in\mathbb{R}^4$ such that $F(x)=0$.
Note that $F(x)$ is a quadratic polynomial in $x$ taking values in $\mathbb{R}^3$.

When $|x|$ very large, $F(x)/|x|^2$ is close to $G(x) = (V(x)/|x|)^*C_0(V(x)/|x|)$, a vector whose norm is $|C_0|$. Since $C$ has rank 1, $C_0$ is not zero. Thus, when $|x|>>0$, $\bigl|F(x)\bigr|/|x|^2\approx |C_0|>0$. Moreover, when restricted to the $3$-sphere of radius $R>0$ in $\mathbb{R}^4$, the homogeneous map $G: S^3(R)\to S^2(|C_0|)$ is simply the Hopf map $S^3\to S^2$, appropriately scaled, which is known not to be homotopically trivial. (In fact, it is a generator of $\pi_3(S^2)\simeq\mathbb{Z}$.)

If there were not a $y\in\mathbb{R}^4$ that satisfied $F(y)=0$, then, for $R>>0$ the map $H(x) = F(x)/|F(x)|$ would map the $4$-ball of radius $R$ to $S^2$ in such a way that the map on its boundary $3$-sphere would be homotopic to the Hopf map, which is impossible.

Thus, there is a $y\in\mathbb{R}^4$ such that $F(y)=0$, i.e.,
$$
A +BV(y) + V(y)^*B^* + V(y)^*CV(y) = \mu\,I_2
$$
for some number $\mu$, as desired.