Things are perhaps a bit messier than you hope. In particular it is not true that the unitary group is non-quasi-split if and only if $v$ ramifies. Disclaimer: I did not know the answer to this question off the top of my head but I did want to know, so I just figured it out below; hopefully there are no errors (hopefully an expert will glance over it and let me know if there are).
From the way you write (talking about "the only one non-quasi-split unitary group...") you seem to be assuming that $v$ is a finite place of $F$ (of course the case of infinite places is very well-known). As you know, if $v$ splits in $E$ then $E\otimes_F F_v$ is isomorphic to $F_v\oplus F_v$ and the unitary group becomes $GL_n$ at $v$. If $v$ does not split (i.e. we are in the inert or ramified case) then there is one prime $w$ above $v$ in $E$, and we have a local extension $E_w/F_v$ of degree 2. So we need to understand unitary groups over local fields in order to answer your question.
So now let $L/K$ be a degree 2 extension of $p$-adic fields and say $V$ is a vector space over $L$ equipped with a Hermitian form (Hermitian for the action of $Gal(L/K)$ of course). If we choose an $L$-basis for $V$ then this form gives rise to a Hermitian matrix $\Phi$ (so $\overline{\Phi}^t=\Phi$). The determinant of $\Phi$ is an element of $L^\times$ which is equal to its Galois conjugate, so it's in $K^\times$. Let $c(\Phi)$ be the image of this determinant in the group $K^\times/N_{L/K}(L^\times)$, a group of order 2. It turns out that this invariant $c$ parametrises isomorphism classes of Hermitian matrices -- so in particular there are two isomorphism classes of Hermitian forms on $V$. For each isomorphism class we get a unitary group.
The next step depends on whether $n$ is odd or even. If $n$ is odd then it turns out that even though the two forms are not isomorphic, the unitary groups are (this can be easily seen -- scaling the form by an element of $K^\times$ can change the isomorphism class of the form but clearly doesn't change the corresponding unitary group). But you are interested in the case $n$ even, and in this case it turns out we get two unitary groups, one quasi-split and one not quasi-split. In particular we see that it is easy to build a non-quasi-split unitary group over $K$ even if $L/K$ is unramified, and it is easy to build a quasi-split unitary group over $K$ even if $L/K$ is ramified -- we just need to get the determinants right. Moreover, we can do all these things over $E/F$ as well, and this is why life is not as simple as you think.
To understand things better and to see what is true, we next need to understand when our local unitary group is quasi-split. You have some global Hermitian form giving rise to a global Hermitian matrix whose determinant is $d\in E$, and what we know so far is that if $v$ is a finite place of $F$ which is not split, and $w$ the unique place of $E$ above $v$, then whether or not $U$ is quasi-split at $v$ depends only on what the image of $d$ is in $F_v^\times/N_{E_w/F_v}(E_w^\times)$. So to see exactly what is going on, I need to tell you which element of $F_v^\times/N_{E_w/F_v}(E_w^\times)$ corresponds to the quasi-split case, and then we also need to check (for our own sanity) that in the global situation $d$ will give us a quasi-split local unitary group for all but finitely many $v$.
Now here's the bad news. It turns out that the story locally (if I worked it out correctly) is the following. We can write $L=K(\sqrt{k})$ for some $k\in K$, and if my calculations are correct, the element of $K^\times/N_{L/K}(L^\times)$ corresponding to the quasi-split unitary group is $k^m$, where $n=2m$. This is because (if I got it right) if we let our Hermitian form be anti-diagonal with entries $+\sqrt{k},-\sqrt{k},+\sqrt{k},\ldots$ (this is Hermitian if I got it right) then this form gives us a quasi-split unitary group because the upper triangular matrices are a Borel. In particular, the naive guess that the quasi-split group corresponds to the identity element of $K^\times/N_{L/K}(L^\times)$ is not always true. The norm of the element $\sqrt{k}\in L$ is $-k$, so it seems to me that the element of $K^\times/N_{L/K}(L^\times)$ corresponding to the quasi-split unitary group is $(-1)^m$, which of course is the identity element iff $(-1)^m$ is the norm of an element of $L$. This happens if $m$ is even or if $L/K$ is unramified (because then any unit is a norm) but not in general.
However, going back to the global situation, we have a global determinant $d\in F^\times$ and at all but finitely many places $d$ will be a local unit, and at all but finitely many places $E_w/F_v$ will be unramified, so it is true that the global unitary group is quasi-split at all but finitely many local places.
In general then, to figure out which places your unitary group is not quasi-split you need to figure out what the determinant $d$ of a corresponding Hermitian form is, and then for each place of $F$ which is either ramified in $E$ or for which this determinant is not a unit (there are only finitely many of these), you need to figure out whether $(-1)^md$ is a local norm or not; the places for which this number is not a local norm are the places where the unitary group is ramified.
