The answer is "yes". Consider $\mathbb S^1$ as the quotient $\mathbb R/\mathbb Z$. Your homeomorphism $f$ lifts to a homeomorphism $\phi : \mathbb R \to \mathbb R$ such that $\phi(x+1)=\phi(x)+1$. Form the map 
$$
h:=\frac{1}{q} \sum _{n=1} ^q (\phi^{\circ n}-pn),
$$ where $\phi ^{\circ n}$ is the composition $n$ times of $\phi$ with itself. By construction 
$$
h\circ \phi = h+\frac{p}{q}
$$
and 
$$
h(x+1)=1+h(x),
$$ so that $h$ factors as a homeomorphism of the circle conjugating $f$ to the rotation.
  By the way this approach works in $\mathbb R^n$ too.