This didn't seem likely, so I looked for a counterexample and found one after a short search. Clearly $Q$ must not have any normal Sylow subgroups, so the first example to try is $Q=S_4$. Then $N$ must not be abelian, so I tried $N=Q_8$ and found an example.
The example is $\mathtt{SmallGroup}(192,988)$ in GAP or Magma. This has seven generators $x_1,\ldots,x_7$, and the normal subgroup isomorphic to $Q_8$ is $\langle x_2, x_3x_7 \rangle$.
Checking that the extension is nonsplit but it splits over the normalizers of Sylow $2$- and $3$-subgroups is routine in GAP or Magma, but I can tell you the commands if it would help.
Magma code follows. Note that the Sylow $2$-subgroup is self-normalizing in $G$. The normalizer of the Sylow $3$-subgroup of $G$ contain $N$, so it is equal to the inverse image of the normalizer in $Q = G/N$ of a Sylow $3$-subgroup.
> G := SmallGroup(192,988);
> N := sub< G | G.2, G.3*G.7 >;
> #N;
8
> IsNormal(G,N);
true
> #Complements(G,N);
0
> P := Sylow(G,2);
> P eq Normalizer(G,P);
true
> #Complements(P,N);
4
> NQ := Normalizer(G, Sylow(G,3));
> N subset NQ;
true
> #Complements(NQ,N);
1