I'll answer this rather indirectly. Let $\pi:SU(4)\rightarrow SO(6)$ denote the double covering map. I'll use the notation $U$ to denote the subgroup $\pi^{-1}(U(3))$ where $U(3)\subseteq SO(6)$ is the one you identified previously.
Proposition 1: The homogeneous space $SO(6)/U(3)$ is naturally diffeomorphic to the homogeneous space $SU(4)/U$.
Proof: Consider the map $\psi:SU(4)/U\rightarrow SO(6)/U(3)$ given by $\psi(g \, U) = \pi(g) \, U(3)$.
The map $\psi$ is well defined: for $h\in U$, we have $\psi(gh\, U) = \pi(gh) \, U(3) = \pi(g)\pi(h) U(3) = \pi(g) U(3) = \psi(g\, U)$ since $\pi(h)\in U(3)$.
The map $\psi$ is surjective: for $g\, U(3) \in SO(6)/U(3)$, pick $h\in SU(4)$ with $\pi(h) = g$. Then $\psi(h\, U) = g\, U(3)$.
The map $\psi$ is injective: assume $\psi(g \, U) =\psi(h\, U)$ for some $g,h\in SU(4)$. Then $\pi(g) \, U(3) = \pi(h)\, U(3)$, so $\pi(g)(\pi(h))^{-1} = \pi(gh^{-1})\in U(3)$. It follows that $gh^{-1}\in \pi^{-1}(U(3)) = U$, so $g\, U = h\, U$.
The map $\psi$ is smooth: this follows from the fact that $SU(4)\rightarrow SU(4)/U$ is a submersion and the fact that $\pi$ is smooth.
The map $\psi^{-1}$ is smooth: To see this, we show that $\psi$ has full rank everywhere, and then appeal to the inverse function theorem. The map $SU(4)\rightarrow SO(6)$ is a submersion because its a covering. Since $SO(6)\rightarrow SO(6)/U(3)$ is a submersion, the composition $SU(4)\rightarrow SO(6)\rightarrow SO(6)/U(3)$ is a submersion.
On the other hand, this composition is the same as the composition $SU(4)\rightarrow SU(4)/U\xrightarrow{\psi} SO(6)/U(3)$, so this composition is a submersion. It follows from the chain rule that $\psi$ must be surjective on tangent spacces.
This completes the proof that $\psi$ is a diffeomorphism. $\square$
(Of course, Proposition 1 is much more general - it works for any covering $G\rightarrow G'$ and subgroup $K\subseteq G'$.)
So we have reduced our problem to identifying $U = \pi^{-1}(U(3))$.
Proposition 2: $U$ is connected.
Proof: We first note that $\pi^{-1}(I) = \{\pm I\}$, so it is enough to find a path in $U$ which connects $I$ with $-I$.
Consider the path $\gamma:[0,2\pi]\rightarrow U(3)$ given by $\gamma(t) = \operatorname{diag}(e^{it},1,1)$, which generates $\pi_1(U(3))$. Under the inclusion $U(3)\rightarrow SO(6)$, this maps to a matrix having a block form $\begin{bmatrix} \cos\theta & \sin \theta \\ -\sin \theta & \cos\theta \end{bmatrix}$ with the other blocks being $0$ or the identity. A matrix of this form generates $\pi_1(SO(6))$, so the loop $\gamma$ is homotopically non-trivial.
Now, lift $\gamma$ to a path $\hat{\gamma}$ in $SU(4)$ where $\hat{\gamma}(0) = I$. If $\hat{\gamma}(2\pi) = I$, then the closed loop $\hat{\gamma}$, which is null homotopic, projects to a null homotopic loop in $SO(6)$. But it projects to the image of $\gamma$, which we've already argued is an essential loop. Thus, $\hat{\gamma}(2\pi) = -I$.
Finally, just note that $\pi(\hat{\gamma}) = \gamma\subseteq U(3)$, so $\hat{\gamma}\subseteq U$. $\square$
So, $U$ is a connected subgroup of $SU(4)$. Further, $\pi$, when restricted to $U$, is a covering of $U(3)$. It follows that $U$ is full rank subgroup of $SU(4)$ and that is has dimension $9$.
One can now turn, for example, to the Borel - de Siebenthal classification of maximal connected max rank subgroups of simple Lie groups or use some simpler representation theory to see the following:
Proposition 3: The only maximal connected max rank subgroups of $SU(4)$ are, up to conjugacy, $U(3)$ and $S(U(2)\times U(2))$.
Since $U$ is connected, $U$ must be a subgroup of one of these. Since $\dim U = 9$, it must be a subgroup of $U(3)$. Since $U(3)$ has dimension $9$, $U = U(3)$.
