Let $S \subseteq \omega_1$ be a stationary set of limit ordinals and let $L = \langle L_\alpha \;|\; \alpha\in S\rangle$ be a ladder system. We say that $L$ has $\kappa$-uniformization if for every sequence of functions $f_\alpha : L_\alpha\to \kappa$ ($\alpha\in S$), there is a function $F : \omega_1\to\kappa$ such that for every $\alpha\in S$, $F\upharpoonright L_\alpha =^* f_\alpha$, i.e. $F$ and $f_\alpha$ agree on a cofinite subset of $L_\alpha$.
It is known (see Eklof-Mekler-Shelah, Theorem 15) that, if $S$ is stationary and every ladder system on $S$ has $2$-uniformization, then in fact every ladder system on $S$ has $\omega$-uniformization.
Question: Suppose $S$ is stationary and $L$ is a ladder system on $S$ which has $2$-uniformization. Does it follow that $L$ has $\omega$-uniformization?
Edit: Since it seems the answer to the above question might be a little far off at the moment, let me ask a weaker question that I'm also interested in:
Question: Suppose $S$ is stationary and $L$ is a ladder system on $S$ which has $2$-uniformization. Let $f_\alpha : L_\alpha\to \omega$ be the collapse map. Does there exist an $F : \omega_1\to \omega$ such that for all $\alpha\in S$, $F\upharpoonright L_\alpha =^* f_\alpha$?
Note that, given $\diamondsuit(S)$, there exists a ladder system on $S$ such that the above "canonical" $\omega$-coloring (i.e. the coloring given by the collapse maps) has no uniformization; hence if the answer is yes, then it must come somehow from the $2$-uniformization property of the ladder system.