Suppose $B\in\mathbb{R}^{m\times n}$ is an $\varepsilon$-spectral approximation to $A\in\mathbb{R}^{m\times n}$, that is, for all $x\in\mathbb{R}^n$, $$(1-\varepsilon)||Ax||_2^2\le ||Bx||_2^2\le (1+\varepsilon)||Ax||_2^2.$$
Is it possible to derive a statement of the form: for all matrices $X\in\mathbb{R}^{n\times m}$, $\mathsf{Tr}(AX)$ and $\mathsf{Tr}(BX)$ are close? Vaguely, I would guess that spectral approximation would imply that $AX$ and $BX$ are close in Frobenius norm, and then perhaps that can be used to show that their traces cannot be too far apart?
Any help is appreciated!
$\operatorname{tr}((A-B)X)$ is not necessarily small when $\|X\|_F$ is bounded and $\varepsilon$ is small. E.g. when $m=n$ and $Q\ne I$ is real orthogonal, we can take $\varepsilon=0$ in your inequality, but $\operatorname{tr}((I-Q)X)$ can be significantly different from zero for some $X$.