This is not an answer; but Margulis and Soifer construct maximal subgroups of lattices $\Gamma$ . Contrary to what Misha says, they are not free products (nor necessarily free groups), but they are maximal subgroups which CONTAIN free subgroups which map onto every finite quotient of the lattice $\Gamma$. It is entirely possible that maximal subgroups are simple (I do not know a proof, one way or another).
Aakumadula
$\DeclareMathOperator\SU{SU}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Cl{Cl}$Yes the the above is the correct list of maximal closed subgroups of $ \SU_3 $.
Antoneli, Forger, and Gaviria - Maximal Subgroups of Compact Lie Groups
classifies all maximal closed subgroups of $ \SU_n $ whose identity component is not simple (here trivial counts as simple). According to this paper, pages 1013–1018, the maximal closed subgroups of $ \SU_3 $ of this type are the normalizer of the maximal torus
$$
N(T)=S(U_1 \times U_1 \times U_1): S_3
$$
as well as
$$
S(U_2 \times U_1 )\cong U_2.
$$
The maximal closed subgroups with trivial identity component are the finite groups:
$$
3.A_6
$$
of order $ 3(360)=1080 $ (known as the Valentiner group)
and
$$
\langle \zeta_3I\rangle \times \GL_3(\mathbb{F}_2)
$$
of order $ 3(168)=504 $. Both these two groups are central extensions by $\langle\zeta_3 I\rangle$ of a finite simple group. But the first is perfect central extension, indeed a Schur cover. While the second is just a direct product. A third finite maximal closed subgroup is the complex reflection group with Shephard–Todd number 25 (see Complex reflection group) and order $ 3(216)=648 $ which happens to be a central extension again by $\langle\zeta_3 I\rangle$ of the Hessian group of order 216. (Since you are from quantum computing this last subgroup would be known to you as the (determinant-1 subgroup of the) qutrit Clifford group.)
The only maximal closed subgroup of $ \SU_3 $ with nontrivial simple identity component is the direct product
$$
\langle\zeta_3I\rangle \times \SO_3(\mathbb{R}).
$$
To summarize, the maximal closed subgroups of $ \SU_3 $ are
\begin{align*}
&N(T)\\
& S(U_2\times U_1)\cong U_2\\
& \langle\zeta_3 I\rangle \times \SO_3(\mathbb{R}) \\
& \langle\zeta_3 I\rangle \times \GL_3(\mathbb{F}_2) = 3 \times PerfectGroup(168) \\
& 3.A_6=PerfectGroup(1080)\\
& S(\Cl_1(3))=SmallGroup(648,532)
\end{align*}
where $ S(\Cl_1(3)) $ is the determinant-1 subgroup of the single qutrit Clifford group. Every closed subgroup of $ \SU_3 $ is contained in one of these $ 6 $ maximal groups.
Also since unitary $ t $ designs are popular in quantum computing it may be of interest to you that $ 3.A_6 $ is a unitary 3-design and $ \zeta_3 \times \GL_3(\mathbb{F}_2) $ and $ S(\Cl_1(3)) $ are both unitary 2-designs.
This is consistent with claim 3 of my answer to Finite maximal closed subgroup of connected Lie group
that all maximal $ 2 $-design subgroups of $ \SU_n $ (all $ 3 $ designs are also $ 2 $ designs) are finite maximal closed subgroups of $ \SU_n $.
The only other closed subgroups of $ \SU_3 $ which are unitary $ t $-designs for $ t \geq 2 $ are the $ \GL_3(\mathbb{F}_2) $ subgroup of $ \langle\zeta_3 I\rangle \times \GL_3(\mathbb{F}_2) $ and the commutator subgroup of the qutrit Clifford group, SmallGroup(216,88), which has size $ 3(72)=216 $. These are again unitary $ t $-designs for $ t=2 $.
In addition to all 5 of these 2-designs there is one other Lie primitive subgroup (i.e. not contained in any proper positive dimensional closed subgroup): it is a subgroup of the qutrit Clifford group of size $ 6(36)=216 $.
Best Answer
The group $(\mathbb{Z}/2\mathbb{Z}) \wr S_{n}$ is finite maximal for $n$ sufficiently large (probably $n>72$ will do, by a theorem of M.Collins on a sharp form of Jordan's theorem on finite complex linear groups, and this is likely more generous than the true bound). I mean here the group of $n \times n$ monomial matrices with all non-zero entries $\pm 1$.
To roughly sketch out the argument for large enough $n$: think of the hyperoctahedral group as a complex linear group of degree $n$. If it is contained in larger finite subgroup X of ${\rm O}_{n}(\mathbb{R}),$ the (viewed as complex linear group), $X$ has an Abelian normal subgroup $A$ with $[X:A] \leq (n+1)!,$ and the only way any index greater than $n!$ (but less than or equal to $(n+1)!$ can be attained is if $X/A$ has a composition factor $A_{n+1}.$ But if $X/A$ has a composition factor $A_{n+1},$ then $X$ must be (irreducible and) primitive as complex linear group, in which case $A$ must consist of scalars. But since $X$ is really a subgroup of ${\mathbb{O}}_{n}(\mathbb{R})$, we have $|Z(X)| \leq 2$ and $|X| \leq (2n+2)n!,$ contrary to $|X| > 2^{n}n!.$ I should add that if $[X:A] \leq n!,$ we get $|A| \leq 2^{n},$ which is also a contradiction.
Later edit: as to the existence or otherwise of other finite maximal subgroups of ${\rm O}_{n}(\mathbb{R}),$ it is true that every non-Abelian finite simple group $G$ is isomorphic to an absolutely irreducible subgroup of ${\rm O}_{n}(\mathbb{R})$ for some $n$ (equivalently, every such simple group has a non-trivial irreducible complex character with Frobenius-Schur indicator $1$). We could consider (necessarily faithful) irreducible character of least degree, say $m$, with this property for $G$.
In this embedding, it might be that the representation extends (as real representation) to some part of the automorphism group of the simple group (that is, to a larger almost simple group with the same unique minimal normal subgroup)). In "most" cases, I would expect that the largest almost simple group to which it extends would be a finite maximal subgroup of ${\rm O}_{m}(\mathbb{R}).$