A much more general fact is true: any isometry of any closed negatively curved Riemannian manifold is not homotopic to the identity. There are many proofs of this; one (perhaps not the most natural) is as follows. Hartman proved that if two harmonic maps $f_0,f_1\colon M\to N$ are homotopic, where $M$ is compact and $N$ is nonpositively curved, then they $f_0$ and $f_1$ are the boundary of an isometrically embedded product $f\colon M\times [0,1]\to N$. If $M=N$ is negatively curved, this is impossible (such a product region would have sectional curvature 0 in certain directions), so we conclude that no nontrivial isometry is homotopic to the identity.
To apply this to your example, note that by uniformization the universal cover of the Riemann surface $C$ is the unit disk $\Delta$. Thus your curve $C$ is the quotient $\Delta/\Gamma$ by some group $\Gamma$ of biholomorphic automorphisms of $\Delta$, namely Aut($\Delta$)=PSL$_2(\mathbb{R})$. Now note that the action of PSL$_2(\mathbb{R})$ on $\Delta$ preserves the Poincaré metric (of constant curvature $-1$), which thus descends to a metric of constant negative curvature on $C$. The resulting metric is preserved by any automorphism of $C$ as a Riemann surface, so in particular your original map is an isometry in this metric.
