The result is true for Hall subgroups in solvable groups, but not in general. I don't have the references to hand, but it's a Theorem of Brauer, or maybe E.C. Dade, or maybe Suzuki, that if $G$ has a Hall $\pi$-subgroup $H$ which controls the fusion of its elements AND every Brauer elementary $\pi$-subgroup of $G$ is conjugate to a subgroup of $H$, then $G$ has a normal $\pi$-complement ( the conditions are also necessary). By Hall's Theorems, the condition on Brauer elementary Subgroups is automatically satisfied if $G$ is solvable ( probably $\pi$-solvable will do).
The proof is as follows. Take an irreducible character $\chi$ of $H$: extend it to a well-defined class-function $\chi^{\ast}$ of $G$ as follows: let $g_{\pi}$ be the $\pi$-part of $g$, and set $\chi^{\ast}(g) = \chi(h),$ where $h$ is any element of $H$ which is conjugate to $g_{\pi}.$ Then by Brauer's characterization of characters, $\chi^{\ast}$ is a generalized character of $G$, using the second assumption. The second condition also ensures (via a counting argument) that $\chi^{\ast}$ is irreducible. Then the normal complement is $\bigcap_{\chi \in {\rm Irr}(H)} {\rm ker}(\chi^{\ast}).$ I think there is a proof in W. Feit's book "Characters of Finite Groups" if more detail is required.
However, if $p$ is a prime greater than $3,$ note that the symmetric group $S_{p}$ has a Hall $p^{\prime}$-subgroup ( namely $S_{p-1}$) which controls the fusion of its elements ( for two elements are conjugate if and only if they have the same cycle type), but there is evidently no normal complement. It is easy to check that there are $p$-regular elements of $S_{p}$ which are not conjugate to any element of $S_{p-1}$ ( choose a product of disjoint $2$- and $3$-cycles which moves all points, for example), so the condition on Brauer elementary $p^{\prime}$-subgroups is not satisfied.