The answer is yes, if $\alpha$ is continuous. Let $g(t):=t\alpha(t)$. Then $0\leq g(t)<t$ for $t\geq 0$.

Lemma. Iterates $g^n$ of $g$ tend to $0$ as $n\to\infty$, uniformly
on $[0,t_0]$ for every $t_0>0$.

Proof of the lemma. The sequence $t_{n+1}=g(t_n)$ starting from  $t_00$ is decreasing and non-negative. So it must have
a non-negative limit. This limit must be a fixed point of $g$, but $g$ has no positive fixed point.

Proof of the theorem. Start from any point $x$. Let us prove that $f^n(x)$ is
a Cauchy sequence. Suppose $m\geq n$. Then $|f^m(x)-f^n(x)|\leq g^n(|f^{m-n}(x)-x|)$. By the lemma this tends to $0$ when $n\to\infty$.

Here I denoted the distance by $|x-y|$. It is a convenient notation even if $x-y$ is not defined.

I guess that the answer is no, if $\alpha$ does not have a continuous majorant
which is strictly less than $1$.