Fix a prime number $p$. Is there a sequence $\{a_k\}_{k \in \mathbb{N}}$ of real numbers with $$\lim_{k \to \infty} a_k = 0$$
such that for any finite $p$-group $G$, and any subgroup $H \leq G$ with $[G : H] \geq k$ we have that $$\frac{|\bigcup_{g \in G} gHg^{-1}|}{|G|} \leq a_k ?$$
I think the answer is no for all $p$.
For $p=2$, let $G$ be a dihedral group, and $H$ a noncentral subgroup of order $2$. Then the the fraction in question is about $1/4$.
For odd $p$ and $k>0$, let $G_k$ be the metacyclic group $$\langle a,b \mid a^{p^{k+1}}=1, b^{p^k}=1, a^b=a^{1+p}\rangle,$$ and $H = \langle b \rangle$. Now there are $p^k-p^{k-1}$ elements of order $p^k$ in $H$, and their centralizers in $G_k$ are equal to the subgroup $\langle a^{p^k},b \rangle$ of order $p^{k+1}$ and index $p^k$ in $G_k$, so the union of their conjugates has order at least $(p^k-p^{k-1})p^k$
Hence the proportion of elements of $G_k$ in the union of the conjugates of $H$ is at least $(p-1)/p^2$, and the index $|G_k:H| = p^{k+1}$ is arbitrarily large.
