No. Indeed $\mathrm{SO}(4)$ satisfies the following condition (which is a first-order existential formula) but not $\mathrm{SU}(3)$:
$$\exists w,x,y,z: [x,w]\neq 1\neq [y,z],\; [w,y]=[w,z]=[x,y]=[x,z]$$
(it just says there are two commuting pairs of non-abelian subgroups).
Indeed $\mathrm{SO}(4)$ even contains a copy of the direct product of two non-abelian free groups.
Let's show this doesn't exist in $\mathrm{U}(3)$. Suppose by contradiction we have such a 4-tuple. Since the $\mathbf{C}$-subalgebra generated by $w,x$ is semisimple and non-commutative, after conjugation it is either $\mathrm{M}_3(\mathbf{C})$ or the block-diagonal $\mathrm{M}_2(\mathbf{C})\times \mathrm{M}_1(\mathbf{C})$. So its centralizer is commutative in both cases, contradiction.
In particular there is no injective homomorphism $\mathrm{SO}(4)\to\mathrm{U}(3)$ (no continuity required).
Here's an alternative argument to show that the centralizer of every non-solvable subgroup $G$ in $\mathrm{GL}_3(C)$, with $C$ any field, is abelian. We can suppose that $C$ is algebraically closed. Indeed, choose a triangulation by blocks of $G$ with irreducible diagonal blocks. If it's trivial (i.e., $G$ is irreducible), the centralizer is reduced to scalars. So it's $1+2$ or $2+1$ (if it were $1+1+1$, $G$ would be solvable). In both case, the subalgebra generated by $G$ contains a diagonal copy of $\mathrm{M}_2(C)\oplus \mathrm{M}_1(C)$. So its centralizer is commutative.