No, GAP has found the following counter-example.

gap> G:=TransitiveGroup(16,9);
gap> H:=Stabilizer(G,1);

$G$ and $H$ are order $32$ and $2$.

gap> R:=IrreducibleRepresentations(G);
gap> U:=R[9];
gap> V:=R[11];

$U$ and $V$ are degree $2$.

Now $G_{(U^H)}$ and $G_{(V^H)}$ are order $8$ and $4$, and $G_{(U^H)} \cap G_{(V^H)} = H$.

We checked that $\forall W$ irr. then $(G_{(W^H)} \cap G_{(V^H)},G_{(U^H)} \cap G_{(W^H)}) \neq (H,H)$.
So we get a counter-example (because in general $H \subseteq G_{(X^H)} \subseteq G$).

No, GAP has found the following counter-example.

gap> G:=TransitiveGroup(16,39);
gap> H:=Stabilizer(G,1);

$G$ and $H$ are order $32$ and $2$.

gap> R:=IrreducibleRepresentations(G);
gap> U:=R[9];
gap> V:=R[11];

$U$ and $V$ are degree $2$.

Now $G_{(U^H)}$ and $G_{(V^H)}$ are order $8$ and $4$, and $G_{(U^H)} \cap G_{(V^H)} = H$.

We checked that $\forall W$ irr. then $(G_{(W^H)} \cap G_{(V^H)},G_{(U^H)} \cap G_{(W^H)}) \neq (H,H)$.
So we get a counter-example (because in general $H \subseteq G_{(X^H)} \subseteq G$).