No. Gaschütz (1953) contains a wealth of information on the Frattini subgroup, including Satz 11 which says that $\Phi(H)$ is “nearly” abelian, in that it cannot have any serious inner automorphisms:
If $H$ is a finite group with $G \unlhd H$ and $G \leq \Phi(H)$, then $\operatorname{Inn(G)} \leq \Phi(\operatorname{Aut}(G))$.
This answers your question:
For $G=D_8$, $I=K_4 \not\leq \Phi(A) = \Phi(D_8) = C_2$. For $G=Q_8$, $I=K_4 \not\leq \Phi(A) = \Phi(S_4) = 1$.
This is actually exactly the criteria for a group $G$ to be the Frattini subgroup of any group, as shown in Eick (1997). However, there are some interesting results along the way:
Stitzing (1970) includes the theorem that if $G$ has a center of prime order and an abelian characteristic subgroup contained in the second center but properly containing the center, then $N$ cannot even be a $H$-normal subgroup of a Frattini subgroup of any group $H$. The dihedral group of order 8 satisfies this, as does the extra-special group of exponent $p^2$ and order $p^3$ for any odd prime $p$.
Hill-Parker (1973) show that if $G$ has size $p^n$ and nilpotency class greater than $n/2$, then $G$ is not an $H$-normal subgroup of a Frattini subgroup of any finite group $H$. In particular, amongst the groups of order $p^3$, only the abelian groups occur as normal subgroups of Frattini subgroups.
van der Waall (1974) handles all groups of order $p^4$. van der Waall–de Nijs (1995) handle order 32. Groups with cyclic centers were handled in Makan (1975).
Some people considered restricting $H$ more severely, for instance Hobby (1960) showed that some subgroups that cannot be derived subgroups of $p$-groups also cannot be Frattini subgroups of $p$-groups. This is generalized in Bechtell (1966) which gives a relativized Satz 11, and determines which groups of order $p^4$ can be Frattini subgroups of $p$-groups.
The “normal subgroup of $H$ contained in the Frattini” language may strike you as weird, and Allenby (1980) showed it is false generality: any such $G$ ($G\unlhd H$, $G \leq \Phi(H)$, $H$ finite) is already the Frattini subgroup of a finite group. This was also shown in Wright (1980).
Eick (1997) is a particularly nice paper in which the converse of Satz 11 is proven, in some sense completely classifying finite groups that occur as Frattini subgroups. In particular she shows that $$\{ \Phi(H) : H, \Phi(H) \text{ is finite } \} = \{ \Phi(H) : H \text{ is finite } \} \supsetneq \{ \Phi(H) : H \text{ is finite soluble } \} \supsetneq \{ \Phi(H) : H \text{ is finite nilpotent } \}$$
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