Generally, unless a hydroxide is in excess over a weak acid as $\ce{H2O2}$ - and it was said it is not - $\mathrm{pH}$ would be always lower than pH of the hydroxide.
$$\mathrm{pH}=\mathrm{p}K_ \mathrm{a}= \log \frac{[\ce{A-}]}{[\ce{HA}]}$$
Additionally, if the conjugate basis $\ce{A-}$ is eliminated by precipitation, the ratio $ \frac{[\ce{A-}]}{[\ce{HA}]}$ is kept low and so does $\mathrm{pH}$.
Note also the hydrogen peroxide is weakly acidic even without addition of sulphuric acid.