$2$-uniformization versus $\omega$-uniformization of ladder systems 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.
 A: For partial progress, let me argue at least that
$2$-uniformization implies $n$-uniformization for any finite $n$.
The idea will be to guess the binary digits of the desired
function separately, and then put these values together.
Suppose we are given a system of functions $f_\alpha:L_\alpha\to
n$ on the fixed ladder system $L_\alpha$ for $\alpha\in S$, where
$S$ is stationary. Choose $k$ large enough so that $n\leq 2^k$,
and for each $i<k$ let $f_\alpha^i(x)$ be the $i^{th}$ binary
digit of $f_\alpha(x)$. Since $f_\alpha^i:L_\alpha\to 2$, we get
by $2$-uniformization a function $F_i:\omega_1\to 2$ that almost
threads the functions $f_\alpha^i$. Thus, $F_i(\beta)$ correctly provides the $i^{th}$ binary digit of $f_\alpha(\beta)$ for almost all $\beta\in L_\alpha$. Combining these individual digit functions into
one function $F:\omega_1\to 2^k$, let $F(\beta)$ be the
number whose $i^{th}$ binary digit is $F_i(\beta)$, for $i<k$.
Since the functions $F_i$ almost-thread the functions
$f_\alpha^i$, it follows that $F$ eventually has all the right
binary digits as $f_\alpha\upharpoonright L_\alpha$, and so it
almost threads the original functions $f_\alpha$.
I'm not sure if this idea can be used to establish
$\omega$-uniformization, since it isn't sufficiently clear to me
whether we can ensure that the digits stabilize quickly enough. 
A: It seems that the answer is no. 
Barney showed (can be found in Foreman's chapter in the handbook, Chapter 3, the section regarding Uniformization Ideal) that it is consistent that $\mathrm{Unif}_\omega \subsetneq \mathrm{Unif}_2$, where $S\in \mathrm{Unif}_2$ if there exists a ladder system on $S$, such that any 2-coloring of the ladder can be uniformized. He showed these are normal countably complete ideals (so contains all non-stationary sets). Hence any $S \in \mathrm{Unif}_2 - \mathrm{Unif}_\omega$ will be stationary and a counter-example.
