As Mikhail Borovoi explained in a comment, the question reduces to "when is a unitary group over a non-Archimedean local field quasi-split"? The answer does not distinguish between ramified or unramified separable quadratic extensions.

Let $E/F$ be a separable quadratic extension of non-Archimedean local fields. Isomorphism classes of pairs $(V,h)$ where $V$ is a finite-dimensional vector space over $E$ and $h$ is a non-degenerate hermitian form on $V$ are classified by $(\dim V, \mathrm{disc}(h))$ where $\mathrm{disc} (h) = \det H \mod N_{E/F}(E^{\times})$ with $H$ the matrix of $h$ in some basis of $V$ over $E$. This is the analogue of the classification of quadratic forms (see Jacobson, A note on hermitian forms https://projecteuclid.org/euclid.bams/1183502551 ). In positive dimension every discriminant in $F^{\times} / N_{E/F}(E^{\times}) = \mathbb{Z} / 2 \mathbb{Z}$ occurs.

As Mikhail Borovoi pointed out, in odd dimension these two forms $h_1,h_2$ have the same unitary group, in fact $h_2 \simeq \lambda h_1$ where $\lambda \in F^{\times} \smallsetminus N_{E/F}(E^{\times})$.

In even dimension only one of the two forms gives rise to a quasi-split unitary group, i.e. Mikhail Borovoi's counterexample generalizes to arbitrary even dimension (although in even dimension greater than 2 the group is not anisotropic). There are probably several ways to see this. For example you can show that Borel subgroups of $U(V,h)$ defined over $F$ correspond bijectively to flags $V_1 \subset \dots \subset V_n$ where $\dim_E V = 2n$, $\dim_E V_1 = 1$, $\dim_E V_{i+1}/V_i = 1$ for $1 \leq i \leq n-1$, and $V_n$ is totally isotropic. Or you can argue by taking Galois cohomology of the short exact sequence of algebraic groups over $F$ (for the étale topology)
$$ U(1) \rightarrow U(V,h) \rightarrow PSU(V,h) $$
and using $H^2(F, U(1)) = 0$ (a consequence of Tate-Nakayama) and the fact that the only inner form of a quasi-split group that is also quasi-split is the trivial one, which follows from the existence of a pinning defined over $F$.

In fact all of this can be understood in the context of Galois cohomology of reductive groups. For example the classification of hermitian forms is essentially equivalent to the Hasse principle (vanishing of $H^1$ of a special unitary group).

no$z\in E$ with $N(z)=\lambda$. Then the hermitian form $$h(z_1,z_2)=z_1 z_1^\rho-\lambda z_2 z_2^\rho$$ does not represent 0 nontrivially. Indeed, if $h(z_1,z_2)=0$ and $z_2\neq 0$, then $N(z_1/z_2)=\lambda$, contradiction. It follows that the group $G=U(E^2,h)$ is anisotropic and therefore, $G$ is not quasi-split. $\endgroup$ – Mikhail Borovoi Aug 4 '17 at 11:094more comments