I think the answer is yes.
We can assume that $G$ is semisimple. As you have seen already, and which is is clear from the commutation relations, we have $[U, U] \leq \prod_{\alpha \in \Phi^+ - \Delta} U_\alpha$. I guess you could show equality by using the Chevalley commutator relations, although you need to be careful when $p$ is small (e.g. $G$ of type $B_n$ and $p = 2$).
Here is an idea for a different way to see this. The containment $[U, U] \leq \prod_{\alpha \in \Phi^+ - \Delta} U_\alpha$ shows that $$\dim U / [U, U] \geq \dim U / \prod_{\alpha \in \Phi^+ - \Delta} U_\alpha = \operatorname{rank} G$$
So it would be enough to show that $\dim U/[U,U] \leq \operatorname{rank} G$.
For this, a result of Steinberg states that there exists a regular unipotent element $u \in G$, which has the property that $\dim C_G(u) = \operatorname{rank} G$. One can show that $u$ is contained in a unique Borel subgroup $B$ of $G$, and that $C_G(u) = C_B(u) = C_U(u)Z(G)$, where $U$ is the unipotent radical of $B$. Proofs can be found in reference (*) below.
In any case, we have a unipotent element $u \in U$ such that $\dim C_U(u) = \operatorname{rank} G$. The conjugacy class $\operatorname{Cl}_U(u)$ of $u$ in $U$ is contained in the coset $u[U,U]$ (proof: $g^{-1}ug = u[u,g]$), so $$\dim \operatorname{Cl}_U(u) = \dim U - \dim C_U(u) \leq \dim [U,U]$$
giving $\dim U/[U,U] \leq \operatorname{rank} G$.
(*) See Chapter 4, sections 4.1 - 4.5 in
Humphreys, James E.: Conjugacy classes in semisimple algebraic groups. Mathematical Surveys and Monographs, 43. American Mathematical Society, Providence, RI, 1995.
or:
Theorem 3.1 in:
Steinberg, Robert: Regular elements of semisimple algebraic groups. Inst. Hautes Études Sci. Publ. Math. No. 25 1965 49–80.
and Lemma 4.3 in:
Springer, T. A.: Some arithmetical results on semi-simple Lie algebras. Inst. Hautes Études Sci. Publ. Math. No. 30 1966 115–141.