If $f$ and $g$ are two hyperbolic isometries of the hyperbolic space $\mathcal H^n$, we know that $f$ has 2 fixed points on $\partial \mathcal H^n$ and similiarly for $g$.
Is it true that $f$ and $g$ generate a non-abelian free group if $\mathrm{fix}(f)\cap \mathrm{fix}(g) = \varnothing$ ? If so, could you provide a reference for this fact ?
That is not true.
A simple example is the following one: consider the hyperbolic isometry
\begin{equation*} \phi:= \left[ \begin{array}{l l} 4 &0\\ 0 &\frac{1}{4}\\ \end{array} \right] \in\text{PSL}(2,\mathbb{R})=\text{Isom}^+\left(\mathbb{H}^2\right) \end{equation*}
and the finite order elliptic element
\begin{equation*} \rho:= \left[ \begin{array}{l l} \frac{1}{2} &-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} &\frac{1}{2}\\ \end{array} \right] \in\text{PSL}(2,\mathbb{R})=\text{Isom}^+\left(\mathbb{H}^2\right) \end{equation*}
with $\rho^6=1$. The product
\begin{equation*} \psi:=\phi\rho= \left[ \begin{array}{l l} 4 &0\\ 0 &\frac{1}{4}\\ \end{array} \right] \cdot \left[ \begin{array}{l l} \frac{1}{2} &-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} &\frac{1}{2}\\ \end{array} \right] = \left[ \begin{array}{l l} 2 &-2\sqrt{3}\\ \frac{\sqrt{3}}{8} &\frac{1}{8}\\ \end{array} \right] \end{equation*}
is again hyperbolic as $\text{tr}(\phi\rho)=2+\frac{1}{8}>2$. Moreover it is easy to see that its fixed points are disjoint from those of $\phi$.
The subgroup $\Gamma:=\langle\phi,\psi\rangle$ cannot be a non-abelian free group as it contains the torsion element $\rho$.