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 $M_3(\mathbf{C})$ or the block-diagonal $M_2(\mathbf{C})\times M_1(\mathbf{C})$. 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).