I will consider $S^1$ as the one-point compactification of the real line, $S^1={\mathbb R}\cup\{\infty\}$ where the smooth atlas is given by the identity chart on ${\mathbb R}$ and the chart given by the map $x\mapsto x^{-1}$ on ${\mathbb R}\cup\{\infty\}- \{0\}$. The map $f: S^1\to S^1$ given by $f(x)=2x, x\ne \infty$, $f(\infty)=\infty$ is a diffeomorphism.
If you do not like this, just take the standard unit circle in the complex plane and let $f$ be the restriction to $S^1$ of a suitable linear-fractional transformation preserving the unit disk.
Now, suppose that $g$ is any Riemannian metric on $S^1$; take $\beta=1$. The derivative $df_0$ of $f$ at $0$ is again the multiplication by $2$. Hence, $||df_0||=2$ where the norm is taken with respect to the metric $g$. But $1<2$.
Hence, the answer to your question is indeed negative.
Here is a more general argument. Let $M$ be any smooth compact connected manifold and let $f: M\to M$ be any degree 1 smooth map, for instance, the identity map. Take $0<\beta<1$. Then for any Riemannian metric $g$ on $M$ there exists a point $x\in M$ such that $||df_x||\ge 1>\beta$. The reason is that otherwise the volume of $f(M)$ is strictly less than the volume of $M$, which contradicts the degree assumption.
