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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 wors in $\mathbb R^n$ too.

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.

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.

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 wors in $\mathbb R^n$ too.

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 $h:=\frac{1}{q} \sum _{n=0} ^q (\phi^{\circ n}-pn)$, 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.

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.