"Universal" means that the construction steps are independent of the length of the given segment. In the Euclidean plane one can take the diagonal of a square built on it. Without the "universal" the answer is affirmative. However, that is due to the fact that the hyperbolic plane has a natural unit of length, and one takes advantage of comparing it to the given length, which is excluded here.

Define the Schweikart's length as the length of a perpendicular to a line such that the asymptotic parallel to it through the endpoint of the perpendicular makes $\pi/4$ angle with the perpendicular. A construction that goes back to Bolyai (the converse to the better known one, see Will Jagy's paper) allows one to construct a segment of Schweikart's length from "nothing", i.e. from a segment of arbitrary length. So one can proceed as follows: construct a Schweikart's segment, if it is incommensurable with the given one we are done, if it is commensurable then construct a segment incommensurable with the Schweikart's (which is easy).

However, this is clearly not "universal", and in fact is similar in spirit to putting marks on the straightedge and measuring. In Euclidean plane at least marked straightedge is a strictly stronger tool. One can think of a universal construction as implementing a continuous (even real analytic) function $f(l)$ of a given length $l$. Since it is supposed to produce an incommensurable for every positive $l$ we have that $f(l)/l$ never takes rational values, i.e. it is a constant. Therefore, if such $f(l)$ exists it should be $f(l)=\alpha l$ with irrational $\alpha$. This is exactly what happens with the diagonal of the square in the Euclidean plane.

My intuition is that the answer is "no", nontrivial elementary hyperbolic constructions transform lengths nonlinearly, but I am not sure. Is there a hyperbolic straightedge and compass construction that does that?

EDIT: Another way to put it is that I am looking for an algorithm that is guaranteed to halt after finitely many steps with an incommensurable for any input. Or rather proving that, as I suspect, none exists. Comparison to the natural length would involve Euclidean algorithm presumably, to test for commensurability, and that would never halt if the input is in fact incommensurable. By laying off and bisecting one can implement $l\mapsto\frac{m}{2^n}l$ with hyperbolic straightedge and compass, but I suspect that these are the only linear transformations possible. Even trisecting a segment $l\mapsto\frac{1}{3}l$ is impossible, Euclidean trisection relies on the intercept theorem, and hence on the parallel postulate.

The length here refers to the standard assignment via binary fractions as in chapter 4 of Greenberg's text for example, which works uniformly in both hyperbolic and Euclidean geometries, not the specialized multiplicative length of Hartshorne. However, that in the hyperbolic case the lengths themselves do not form (the positive part of) a field, but rather their exponentiations do, is perhaps the reason why no such algorithm exists. Since $x=e^l$ an equivalent question would be if $x\mapsto x^\alpha$ with some irrational $\alpha$ can be done with Euclidean straightedge and compass for any length $x$, probably not, but I don't know how to prove that either.