Answer by Adam Zalcman for Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt bases?

TL;DR: No such relation exists, because the upper bound on the norm fails to impose any constraints whatsoever on the basis elements corresponding to very small Schmidt coefficients $\sqrt{p_i}$ and $\sqrt{q_i}$.

Let's construct an explicit counterexample. Define the following states of two qutrits $A$ and $B$$$\begin{align}|\phi\rangle^{AB}:=\sqrt{1-\delta^2}|0\rangle|0\rangle+\delta|1\rangle|1\rangle\tag1\\|\psi\rangle^{AB}:=\sqrt{1-\delta^2}|0\rangle|0\rangle+\delta|2\rangle|2\rangle\tag2\end{align}$$where $\delta:=\varepsilon/2$.Note that the matrix of $D:=|\phi\rangle\langle\phi|^{AB} - |\psi\rangle\langle\psi|^{AB}$ in the computational basis has six non-zero elements: two diagonal ones equal to $\pm\delta^2$ and four off-diagonal equal to $\pm\delta\sqrt{1-\delta^2}$. Therefore, the Frobenius$^1$ norm of $D$ is$$\begin{align}\|D\|_2&=\sqrt{2\delta^4+4\delta^2(1-\delta^2)}\tag3\\&=\sqrt{4\delta^2-2\delta^4}\tag4\\&\le 2\delta=\varepsilon.\tag5\end{align}$$If $\delta$ is so small that $\sqrt{1-\delta^2}\gt\delta$, then$$\langle e_2|^A\langle \tilde{e}_2|^B|f_2\rangle^A|\tilde{f}_2\rangle^B=\langle 1|^A\langle 1|^B|2\rangle^A|2\rangle^B=0\tag6$$so the only possible $g(\varepsilon)$ is $g(\varepsilon)=1$ which yields the trivial bound $|\langle e_i|^A\langle \tilde{e}_i|^B|f_i\rangle^A|\tilde{f}_i\rangle^B| \geq 0$.

^{$^1$ I assume that $\|.\|_p$ in the question denotes the Schatten $p$-norm, so we get Frobenius norm for $p=2$. Norm equivalence will provide tight bounds on other norms, too. For example, for the trace norm we have $\|D\|_1\leq\varepsilon\sqrt{3}$.}