Here is a tentative (negative) answer to the question of whether the existence of an $E_{0}$ (Vitali) selector implies the existence of a nonprincipal ultrafilter on $\omega$. I need to check some of the details (I'll say where below) but it seems ok to me. I'm posting this now in hope that someone has any ideas for an easier solution.

Let $P$ be the partial order consisting of countable partial $E_{0}$ selectors, ordered by includion. Let (a) be the statement (meant to be applied in the context of Choice) that whenever $D$ is a dense subset of $P$ in $L(\mathbb{R})$ and $F \subseteq P$ is a filter of cardinality less than the continuum ($\mathfrak{c}$), $F$ can be extended to a filter containing a member of $D$. If (a) holds, the continuum is regular, and $|\mathcal{P}(\mathbb{R}) \cap L(\mathbb{R})| = \mathfrak{c}$ (which holds if a measurable cardinal exists, for instance), then one can build $L(\mathbb{R})$-generic filters for $P$.

For all I know, (a) is a consequence of ZFC plus suitable large cardinals. It is clearly a consequence of the Continuum Hypothesis, but that doesn't seem to be of much use. Assuming that the theory of $L(\mathbb{R})$ is fixed by set forcing (which happens for instance if there exist proper class many Woodin cardinals; this hypothesis can be fine-tuned for our purposes), you would get that forcing with $P$ over $L(\mathbb{R})$ does not add a nonprincipal ultrafilter, if you knew that each of the following statements could be forced along with (a) + `$\mathfrak{c}$ is regular" : (1) for every nonprincipal ultrafilter $U$ on $\omega$ there is a Ramsey ultrafilter in $L(\mathbb{R})[U]$; (2) there are no Ramsey ultrafilters. Each of (1) and (2) is forceable along with $\mathfrak{c}= \aleph_{2}$. I'm not sure who showed this; my guess is Blass in the first case, and Kunen in the second.

I don't know anything about producing models of (a) by forcing over models of ZFC, but (a) holds in the $\mathbb{P}\mathrm{max}$ extension of any model of $\mathrm{AD}^{+}$ (such as $L(\mathbb{R})$, if there exist infinitely many Woodin cardinals below a measurable cardinal). This does not seem to give much information on the original question, either, though it shows (via a $\Delta$-system argument) that if $G \subseteq \mathbb{P}\mathrm{max}$ is an $L(\mathbb{R})$-generic filter, then in $L(\mathbb{R})[G]$ there are $L(\mathbb{R})$-generic filters for $P$ whose extensions do not contain any ultrafilter on $\omega$ generated by a tower of length $\omega_{2}$ from the point of view of $L(\mathbb{R})[G]$.

It seems to me that there are natural variations of $\mathbb{P}\mathrm{max}$ for producing models in which the cardinal characteristic $\mathfrak{u}$ is equal to $\aleph_{1}$ (I believe that Woodin has done this, in fact). In these models, (a) seems to hold, and the cardinal characteristic $\mathfrak{g}$ should be $\aleph_{2}$, which gives statement (1), by results of Laflamme.
More importantly, it gives Near Coherence of Filters (NCF), the statement that any two nonprincipal ultrafilters on $\omega$ are finite-to-one reducible to a common ultrafilter. I don't know about getting (2)+(a) with a $\mathbb{P}\mathrm{max}$ variation. However, if there is a $P$-name for an nonprincipal ultrafilter on $\omega$, then one can map this name to two (continuum many, in fact) independent parts of $P$, and thus show that NCF cannot hold in the $P$-extension of $L(\mathbb{R})$ unless there are no nonprincipal ultrafilters there.

This argument seems to have nothing to do with $E_{0}$. If it is correct, then it should show, for instance, that if $L(\mathbb{R}-sharp) \models \mathrm{AD}^{+}$, then the partial order which adds a choice function for $L(\mathbb{R})\cap \mathcal{P}(\mathbb{R})$ with countable conditions does not add a nonprincipal ultrafilter on $\omega$.

I expect that this argument is more complicated than necessary in more than one way. One would expect the proof to go though a determinacy-style $\Delta$-system lemma, but I don't see how to do that.