Can you find a function $beta(x)$ where if $a+b=n^m$ then $beta(frac{a}{b})$ is irrational?

This is only an 'almost' answer, since the function constructed may fail to be differentiable on some set of measure $0$.

It is known that any monotone increasing real-valued function is differentiable almost everywhere (see, for example, this Wikipedia link or this question on this site). Then it is quite easy to construct a monotone function satisfying your criteria:

Starting with $beta(0) = pi$, choose an enumeration $(q_n)$ of the remaining rationals and, one at a time, pick $beta(q_n)$ to be rational or irrational according to the criteria and so that the resulting partial function is strictly monotone increasing. More precisely, for any $q_n$, there must be a greatest $q_j<q_n$ and a least $q_k>q_n$ such that $j,k<n$, and the interval $(q_j,q_k)$ contains infinitely many rationals and uncountably many irrationals to pick from (if you care about the axiom of choice, there's no need for it here: you can always narrow down your search to some well-orderable subset of the irrationals such as $sqrt{2}mathbb{Q}$).

Then, for irrrational $x$, define $beta(x)$ to be the supremum of $beta(q)$ for all rational $q<x$. The resulting function is (strictly) monotone increasing and is therefore differentiable almost everywhere.