Lemma: If $f,g:S^1\to S^1$ are two orientation-preserving homeomorphisms and $f\circ g=g\circ f$, then $\rho(f\circ g)=\rho(f)+ \rho(g) (\mod 1)$ (This is an exercise from Barreira & Valls' Dynamical Systems: an Introduction, p. 84).
Proof of Lemma: Since $f,g$ are orientation-preserving, there are lifts $F,G:\mathbb{R}\to\mathbb{R}:F \uparrow,G\uparrow$ of $f,g$, respectively. Since $f,g$ commute, so do $F,G$: Let $\pi:\mathbb{R}\to S^1$ be the projection. Then $F\circ G=\pi^{-1}\circ f\circ \pi \circ \pi^{-1} \circ g\circ \pi=\pi^{-1} \circ f\circ g \circ \pi=\pi^{-1}\circ g\circ f\circ \pi=G\circ F$.
$F\circ G$ is a lift of $f\circ g$ and it is increasing. Hence $\rho(F\circ G)$ is defined. Then $\rho(F\circ G)=\lim_{n\to\infty}\dfrac{(F\circ G)^n(x)-x}{n}= \lim_{n\to\infty}\left(\dfrac{F^n(G^n(x))-G^n(x)}{n}+\dfrac{G^n(x)-x}{n}\right)= \rho(F)+\rho(G)$ since the used limits exist independently of $x\in\mathbb{R}$. Also note that in the second equality we use the fact that $F,G$ commute.
Finally since by definition $\rho(f)=\pi(\rho(F))$, modding the acquired equality by $1$, we are done (Admittedly we won't be using the complete form of the lemma, but I figured I should record it completely for future reference.).
Turning back to your question, we have an orientation-preserving homeomorphism $f:S^1\to S^1$. Let $F$ be an increasing lift of $f$. Then we have $f\circ f^{-1}=1_{S^1}=f^{-1}\circ f$ and $1_{\mathbb{R}}$ is a lift of $1_{S^1}$. As $f$ and its inverse commute, we may employ the above lemma. Noting that $\rho(1_{\mathbb{R}})=0$, as (again) the limit in the definition of the rotation number is independent of $x$, the result follows.
The following argument is an extension of the one for a homeomorphism $h$ in [Hasselblatt & Katok. A First Course in Dynamics. Cambridge University Press, 2003, pg. 128].
We will show (for the right choice of $G$), $\frac{|F^n(x) - G^n(x)|}{n} \to 0$ as $n \to \infty$ for all $x$, where $F$ and $G$ are lifts of $f$ and $g$ respectively, i.e. $\pi F = f \pi$, where $\pi: \mathbb{R} \to \mathbb{S}^1$ is the projection operator.
Let $H$ be a lift of $h$. Then $\pi F H = f \pi H = f h \pi= h g \pi = h \pi G = \pi H G$. This implies $F \circ H = H \circ G + k$ for some $k \in \mathbb{Z}$. Another lift $G'$ of $g$ may be chosen so that $F \circ H = H \circ G'$. In particular, let $G'(x) = G(x) + k$. Since $H$ is a degree-one lift, $H(x + k) = H(x) + k$ for any integer $k$. Then $H \circ G'(x) = H (G(x) + k) = H(G(x)) + k = F \circ H(x)$.
For the rest of the argument, assume the lift $G$ is chosen so that $F \circ H = H \circ G$.
By induction $F^n\circ H = H \circ G^n$: this holds true for $n = 1$ and assuming $F^{n-1} \circ H = H \circ G^{n-1}$ then
$\begin{aligned} F^n\circ H &= F^{n-1}\circ F \circ H\\
&= F^{n-1} \circ H \circ G\\
&= H \circ G^{n-1} \circ G\\
&= H \circ G^n \end{aligned}$.
This implies $G$ may be chosen so that $F^n\circ H = H \circ G^n$, where $H$ is any lift of $h$.
Suppose $H$ is the lift such that $H(0) \in [0,1)$. Then for all $x \in [0,1)$, since $0 < H(0) \leq H(x) < H(1) = H(0) + 1 < 2$, we have $ -1 \leq H(x) -x \leq 2$ for all $x \in [0,1)$. Since $H$ is a degree-one lift, $H(x) - x$ is periodic with period $1$, so this implies for all $x \in \mathbb{R}$, $|H(x) - x| < 2$.
Thus, $|H\circ G^n(x) - G^n(x)| < 2$.
Using the fact $F^n(x + 1) = F^n(x) + 1$ for all $x$, we have $F^n(x + 2) = F^n(x) +2$, and $F^n(x-2) = F^n(x) - 2$. Suppose $| y - x | < 2$, so $ x - 2 < y < x + 2$. Since $F^n$ is strictly increasing, $F^n(x-2) < F^n(y) < F^n(x + 2)$, hence $F(x) - 2 < F^n(y) < F^n(x) + 2$. Therefore, if $|y-x| < 2$, we have $|F^n(y) - F^n(x)| < 2$.
Since $|x - H(x)| < 2$, we have $|F^n(x) - F^n\circ H(x)| < 2$.
Therefore,
$\begin{aligned}
|F^n(x) - G^n(x)| & = |F^n(x) - F^n\circ H(x) + F^n\circ H(x) - G^n(x)|\\
&= |F^n(x) - F^n\circ H(x) + H\circ G^n(x) - G^n(x)|\\
&\leq |F^n(x) - F^n\circ H(x)| + |H\circ G^n(x) - G^n(x)|\\
&< 2 + 2
\end{aligned}$
Hence, $\frac{|F^n(x) - G^n(x)|}{n} < \frac{4}{n} \to 0$ as $n \to \infty$.
This implies $\rho(f) = \rho(g)$.
Best Answer
It is not true.
Let the universal covering map $\mathbb{R} \to S^1$ be $t \mapsto e^{2 \pi i t}$.
Let $f : S^1 \to S^1$ fix $1+0i$, and let $f$ act on $S^1 - \{1+0i\}$ by moving each point a little bit in the counterclockwise direction. Lift $f$ to $F$ so that it fixes each integer $n$ and displaces every other point a little bit to the right.
Let $g : S^1 \to S^1$ fix $-1+0i$ and let $g$ act on $S^1 - \{-1+0i\}$ by moving each point a little bit in the counterclockwise direction. Lift $g$ to $G$ so that it fixes each half integer $n+\frac{1}{2}$ and displaces every other point a little bit to the right.
Then $F \circ G$ displaces each point to the right by an amount bounded away from zero and so $\rho(F \circ G) > 0$. But $\rho(F) + \rho(G) = 0 + 0 = 0$.