$\DeclareMathOperator\Res{Res}\DeclareMathOperator\Gal{Gal}$As noted, the answer is yes. Let $G$ be a simple algebraic group (meaning $F$-simple) over $F$.
Assume first that $F$ is Galois over $\mathbb{Q}$. Then $(\Res_{F/\mathbb{Q}}G)_{F}\simeq G_{1}\times\cdots\times G_{r}$, where the $G_{i}$ are the conjugates of $G$ under $\Gal(F/\mathbb{Q})$. If $H$ is a smooth connected normal algebraic subgroup of $\Res_{F/\mathbb{Q}}G$, then $H_{F}$ is a product of a certain number of the $G_{i}$ (Milne, Algebraic Groups, 21.51), which is stable under the Galois action. But $\Gal(F/\mathbb{Q})$ acts transitively on the set of $G_{i}$, and so $H=\Res_{F/\mathbb{Q}}G$.
In the general case, the same argument works if $G$ is geometrically simple. If not, then we may suppose that $G=\Res_{F^{\prime}/F}G^{\prime}$ with $G^{\prime }$ geometrically simple (ibid. 24.3) and apply the argument to $G^{\prime}$.
This argument fails for algebraic groups like $\mathbb{G}_a^n$ because for them the decomposition into simple objects is not unique.
