The problem is as in the title.
Let $A$ and $B$ be $2 \times 2$ complex matrices. Are there $A,B$ such that for any $2 \times 2$ matrix $C$, either $CAC^{-1}$ or $CBC^{-1}$ is never real. A numerical example I think I found would be:
$A=\left( \begin{array}{cc} -1 & -4-i \sqrt{29} \\ \frac{8}{9}+\frac{1}{9} \left(-4-i \sqrt{29}\right) & -1 \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} -1 & -9 \\ 1 & 3 \\ \end{array} \right)$ However, I do not know how to show that it actually works. It is a candidate as Mathematica could not find the conjugation.
Here is a counterexample. Let $A=\pmatrix{1\\ 0}\pmatrix{1&1}$ and $B=\pmatrix{z\\ \overline{z}}\pmatrix{1&1}$ where $z$ has nonzero real and imaginary parts. Note that all of $A,B$ and $AB$ have real spectra. However, since $\operatorname{tr}(AB^2)=z(z+\overline{z})$ is non-real, $A$ can $B$ cannot be both similar to real matrices via the same similarity transform.
(By adding some appropriate real scalar multiples of $I$ to $A$ and $B$, one can create counterexamples with invertible $A$ and $B$ as well, although $AB$ will then have only a real trace but not necessarily a real spectrum.)