Question: Suppose $g$ is $\alpha$-Hölder continuous and $f\_n$ converges uniformly to $f$ at rate $a\_n$, so that

$a\_n \sup\_{x}|f\_n(x) - f(x)| \to 0 $ as $n \to \infty$

Then does $g \circ f\_n$ converges uniformly to $g\circ f$ at rate $a\_n$?

Answer: 
The claim is false in general. To see it, 
take $f\_n(x) = x + \frac 1n$, $f(x) = x$, and set $a\_n = n^{1/2}$ (for instance), let also $g(x)= x^{1/2}$, where $x\geq 0$. 

We have that $g$ is Holder-$\frac 12$, but the claim in your question fails for these data.

Indeed, $f\_n$ converges to $f$ uniformly on $\mathbb{R}$, in particular 
$$
n^{1/2} \sup\_{x \in [0,1]} |f\_n(x) - f(x)| \to 0.
$$
On the other hand,
$$
n^{1/2} \sup\_{x\in [0,1]} |g(f\_n(x)) - g(f(x))| = n^{1/2} \sup\_{x\in [0,1]} \left| \frac{ f\_n(x) - f(x) }{\sqrt{f\_n(x)} + \sqrt{f(x)}} \right| = \\
n^{1/2} \frac 1n \sup\_{x\in [0,1]} \left| \frac{ 1 }{\sqrt{x+1/n} + \sqrt{x}} \right| \geq (\text{ take } x=0) \\ n^{-1/2} \frac{1}{\sqrt{1/n}} = 1,
$$
so the convergence fails.
No. 

We need $a\_n \sup\_x|f\_n(x) - f(x)|^\alpha \to 0$

From the Hölder continuity of $g$, there are non-negative constant $C$, $\alpha$ such that

$ a\_n|g(x) - g(y)| \le a\_n C |x - y |^{\alpha} $

For $\alpha > 1$, the function $g$ is constant and the convergence is automatic. For $\alpha < 1$, we have, from the uniform convergence of $f\_n$ at rate $a\_n$, that for any $\epsilon > 0$, $\alpha \le 1$, $x$, there exists a $N > N\_\*$ such that for $n > N$

$a\_nC|f\_n(x) - f(x)|^\alpha < \epsilon $

Which implies that for any $\epsilon > 0$ and $x$, we have a $n$ so that

$ a\_n|g(f\_n(x)) - g(f(x))| \le a\_n C |f\_n(x) - f(x) |^{\alpha} < \epsilon$

which proves the result.