No.
Here is an example without small Schmidt coefficients.
To this end, consider$$\lvert\phi\rangle = a\lvert0\rangle\lvert0\rangle + b \lvert1\rangle\lvert1\rangle\ ,$$and$$ \lvert\psi\rangle = a\lvert+\rangle\lvert+\rangle + b \lvert-\rangle\lvert-\rangle\ ,$$where $a=\sqrt{\tfrac12-\varepsilon}$, $b=\sqrt{\tfrac12+\varepsilon}$ [and with $\lvert \pm\rangle = \tfrac12(\lvert0\rangle\pm\lvert1\rangle)$].
$\lvert\phi\rangle$ and $\lvert\psi\rangle$ are in their Schmidt decomposition, and it is unique (as long as $\varepsilon\ne 0$). Moreover,$$\|\lvert\phi\rangle\langle\phi\rvert-\lvert\phi\rangle\langle\phi\rvert\|_p \to 0$$as $\varepsilon\to 0$.
Yet, their Schmidt vectors do not become close to each other; in fact, they are completely independent of $\varepsilon$.
Thus, the only way in which this can be made to work is if you insist that you are sufficiently far (as comapred to $\varepsilon$) from a state with degenerate Schmidt coefficients. Note that this is also the issue with the other example by @AdamZalcman: There is one zero Schmidt coefficient and one very small one, which are thus almost degenerate.