It is a very nice question.

I claim that there is a non-determined dense Vitali set $A$, by a modification of the usual construction of a non-determined set.

First, enumerate all the possible strategies for Alice or Bob in a
well-ordered sequence
$\langle\sigma_\alpha\mid\alpha\lt\frak{c}\rangle$ of length
continuum. A strategy is simply a function mapping the finite play so far to the next digit to be played, and there are at most continuum many such functions. Let us adopt the usual notation, by which $x\star\sigma$ means the play resulting from having Alice play the sequence $x$ and Bob play according to the strategy $\sigma$, and similarly for $\sigma\star y$, where Alice plays $\sigma$ and Bob plays $y$.

Next, in a transfinite recursive procedure, we will make
promises about certain sequences being definitely in the payoff set $A$ and other
promises about certain sequences being definitely not in $A$.

At stage $\alpha$, let $A_\alpha$ be the set of sequences that we
have promised definitely to be in the payoff set $A$, and
$B_\alpha$ is the set of sequences that we have promised to be out
of $A$. These will be disjoint sets of size $|\alpha|$, which is
less than the continuum. Consider now the strategy $\sigma_\alpha$,
treating it first as a strategy for Bob. Let $x$ be a play for
Alice that does not agree on a tail with Alice's play in any
sequence of $A_\alpha\cup B_\alpha$. Such an $x$ exists because the
sets are small and there are continuum many possible plays. Let us
place $x*\sigma_\alpha$ into $A$, that is,
$A_{\alpha+1}=A_\alpha\cup\{x*\sigma_\alpha\}$. This defeats
$\sigma_\alpha$ as a winning strategy for Bob. Next, consider
$\sigma_\alpha$ as a strategy for Alice. Let $y$ be a sequence that
does not agree on a tail with any play by Bob for any sequence used
up to this stage, that is, in $A_{\alpha+1}\cup B_\alpha$. We place
$\sigma_\alpha\star y$ into $B$, meaning
$B_{\alpha+1}=B_\alpha\cup\{\sigma_\alpha\star y\}$. This
defeats $\sigma_\alpha$ as a winning strategy for Alice.

Having completed the recursion, let $A=\bigcup_\alpha A_\alpha$,
and $B=\bigcup_\alpha B_\alpha$ be the resulting sets of size
continuum. By construction, $A$ and $B$ are disjoint, and no two
elements of $A\bigcup B$ contain sequences that have only finite
difference, since every time we added a new sequence either to $A$
or to $B$, it was infinitely different from all the earlier
sequences we added. Thus, $A\cup B$ contains at most one element
from each tail equivalence class. The usual argument is completed here, since $A$ is not determined, as every strategy was defeated with respect to it.

But in order to get the full Vitali property, we do a bit more. By Zorn's lemma, we may
extend $A\subset V$ to a maximal such set disjoint from $B$. That is, $V$ contains $A$, is disjoint from $B$ and no two elements agree on a tail. It follows by maximality that $V$ is a Vitali set.
Furthermore, like $A$, the set $V$ is not determined, since by design, any given
strategy $\sigma_\alpha$ has a specific play, chosen at stage
$\alpha$, which defeats it as winning for either Alice or Bob.

Finally, we can easily arrange that $V$ is dense, by modifying the
construction to start with a fixed countable dense set $A_0$
containing tail-inequivalent sequences.

So this is a dense non-determined Vitali set, as desired.