The answer is yes. Take any $x_0$ and define $x_{n+1}=f(x_n)$.
Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$
Therefore the series
$$\sum|x_{n+1}-x_n|$$
is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.

(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)