Weyl's inequalities for singular values give you the desired upper bound and, in some cases, a lower bound.
For the upper bound, we have $$ |\sigma_i(A) - \sigma_i(B)| \leq \sigma_1(B - A) = \|B - A\|, $$ where $\|\cdot\|$ denotes the spectral norm. This can easily be extended to other unitarily invariant norms.
For the lower bound, one situation where a lower bound is guaranteed is where $\sigma_1(A) \leq \sigma_n(B)$. In this case, we note using the variational characterization of singular values that $$ \sigma_1(A) = \max_{\|x\| = 1}\|Ax\| \leq \min_{\|x\| = 1}\|Bx\| = \sigma_n(B) $$ and that $$ \begin{align} \|B - A\| &= \sigma_1(B-A) = \max_{\|x\| = 1}(B - A)x \\ & \geq \max_{\|x\| = 1} (\|Bx\| - \|Ax\|) \\ & \geq \min_{\|x\| = 1}\|Bx\| - \max_{\|x\| = 1}\|Ax\| = \sigma_n(B) - \sigma_1(A), \end{align} $$ so that $\sigma_n(B) - \sigma_1(A) \leq \|B - A\|$$\sigma_n(B) - \sigma_1(A) \geq \|B - A\|$. From there, it follows that for all $i$ $\sigma_n(B) \leq \sigma_i(B)$ and $\sigma_1(A) \geq \sigma_i(A)$, so that $$ \sigma_i(B) - \sigma_i(A) \geq \sigma_n(B) - \sigma_1(A) \geq \|B - A\|. $$
I suspect that this can be extended to other unitarily invariant norms without too much trouble.