In a somewhat different direction from Alireza: the conjecture is true for a large family of groups, including all abelian groups and many supersolvable groups.
Let me start with the abelian case. Pick an element $g$ of highest possible prime-power order in an abelian group $G$. Then $\langle g \rangle$ has a complement: that is, there is a subgroup $K$ such that $K \langle g \rangle = G$ and $K \cap \langle g \rangle = 0$. In particular, $K \cong G / \langle g \rangle $ by the Isomorphism Theorems, and for any subgroup $X$ with $\langle g \rangle \subseteq X$ there is a corresponding subgroup $X \cap K$ which does not contain $\langle g \rangle$.
In fact, the same argument applies to any $G$ and prime-power order element $g$ if 1) $g$ generates a normal subgroup, and 2) we can find a complement $K$ to $\langle g \rangle$ in G. In this situation, $[\langle g \rangle, G] \cong [1,K]$.
(Edit: deleted discussion of supersolvable groups, which is irrelevant in light of update below.)
UPDATE: The conjecture is true for all finite solvable groups.
Proof: Let $G$ be a solvable group. Then $G$ has a normal subgroup $N$ of prime index, and some element of prime-power order $g \notin N$. Since $N$ is maximal in $G$, we have $\langle g,N \rangle = G$, and since $N$ is normal we have $\langle g,N \rangle = \langle g \rangle N$. Then by Dedekind's identity, we get that $\langle g \rangle (H \cap N) = H \cap G = H$ for any $H$ containing $\langle g \rangle$.
The last tells us that the map from the interval $[\langle g \rangle,G] \rightarrow [1,N]$ given by $H \mapsto H \cap N$ is an injection. Since $N$ doesn't contain $g$, we get the conjectured statement. $\square$
Indeed, the above works whenever $G$ has a maximal subgroup of prime index. (E.g., for symmetric groups.)
(Thanks to John Shareshian for several useful comments and discussion.)