It seems that the idea you used for the classical case extends to the quantum case.

Let $H = 1_{[0,\infty)}$. Let $\Pi$ be the projector defined by functional calculus as $\Pi=H((1+\sqrt{\epsilon})\sigma - \rho)$ and $\Pi'=1-\Pi$. Observe that $\mathrm{tr}(\Pi'(\rho- (1+\sqrt{\epsilon})\sigma)) \geq 0$

Now write $$ \epsilon \geq \|\rho-\sigma\|_1 \geq \mathrm{tr} (\Pi'\rho) - \mathrm{tr}(\Pi'\sigma) \geq \sqrt{\epsilon} \ \mathrm{tr} (\Pi' \sigma)$$ so that $\mathrm{tr}(\Pi \sigma) \geq 1- \sqrt{\epsilon}$. By construction, we have $\Pi ((1+ \sqrt{\epsilon})\sigma - \rho) \Pi \geq 0$ and therefore $\Pi \rho \Pi \leq (1+\sqrt{\epsilon})\sigma$.

Edit : the following argument does not answer the question and actually appears already in the OP's linked note.

Let $H = 1_{[0,\infty)}$. Let $\Pi$ be the projector defined by functional calculus as $\Pi=H((1+\sqrt{\epsilon})\sigma - \rho)$ and $\Pi'=1-\Pi$. Observe that $\mathrm{tr}(\Pi'(\rho- (1+\sqrt{\epsilon})\sigma)) \geq 0$

Now write $$ \epsilon \geq \|\rho-\sigma\|_1 \geq \mathrm{tr} (\Pi'\rho) - \mathrm{tr}(\Pi'\sigma) \geq \sqrt{\epsilon} \ \mathrm{tr} (\Pi' \sigma)$$ so that $\mathrm{tr}(\Pi \sigma) \geq 1- \sqrt{\epsilon}$. By construction, we have $\Pi ((1+ \sqrt{\epsilon})\sigma - \rho) \Pi \geq 0$ and therefore $\Pi \rho \Pi \leq (1+\sqrt{\epsilon})\Pi \sigma\Pi$.