Hi,
I just saw a theorem in a paper of Isaacs which could be useful.
Assume additional to the general case(with same notation)that N is normal in H and H/N is abelian.
Then the lemma:                               

"Let N be a normal subgroup of H and H/N abelian.Let $\vartheta \in Irr(H) $and $\psi \in Irr(N)$ and $[\psi,\vartheta_N] \neq 0$.Then every Z $\in Irr(H)$ with $[Z_N,\psi] \neq 0$ has the form $Z=\lambda \vartheta$ for a linear $\lambda \in Lin(H/N) $ " 
            
tells us that every irreducible constituent of $ \psi^{H} $ has the form $\lambda \vartheta$,where $\lambda \in Lin(H/N)$ because if Z is an irreducible constituent of $ \psi^{H} $ then $[Z,\psi^{H}]=[Z_N,\psi]$.We can then write $ \psi^{H} $ as a sum of all different $\lambda \vartheta$ with multiplicity one,because $[\psi^{H} , \lambda \vartheta]=[\psi, (\lambda \vartheta)_N]=[\psi,\psi]=1$.           

We have $[\chi_H , \psi^{H}]=[\chi,\psi^{G}]=[\chi_{1+J^{2}},\psi] \neq 0$
Choose an irreducible constituent of $ \psi^{H} $,called $\phi$ with $[\phi,\chi_H] \neq 0.$
On the one hand we have: $(\psi^{H})_{1+J^{2}} = |H:(1+J^{2})| \psi $ 
and on the other (since $[\phi,\psi^{H}]=[\phi|{1+J^{2}} , \psi]$):    

$(\psi^{H})_{1+J^{2}} = a \phi +etc$ 
Comparing these 2 gives: $\phi|{1+J^{2}}=c \psi$ 
for natural a  and c and etc as a combination of other caracters.
we have to show that c=1 to finish the proof.
$c=[\phi|{1+J^{2}} , \psi]=[\phi,\psi^{H}]$,but we showed above that $\psi^{H}$ has only irreducible constituents with multiplicity one.
I am thankful for proofreading this or giving hint to avoid the additional assumptations.