in your question formulation, you have forgotten to take into account $\ce{H2O2}$ is a weak acid.
Generally, unless $\ce{Ca(OH)2}$ is in excess over a weak acid as $\ce{H2O2}$ - and it was said it was not - $\mathrm{pH}$ would be always lower than pH of the hydroxide.
$$\mathrm{pH}=\mathrm{p}K_ \mathrm{a} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$
The hydroxide forms from $\ce{H2O2}$ the $\mathrm{pH}$ buffer solution of a weak acid and it's salt.
$$\ce{H2O2 <<=> H+ + HO2-}$$ $$\ce{H+ + OH- <=>> H2O}$$
Additionally, if $\ce{HO2-}$ is partially eliminated by precipitation, the ratio $ \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$ is kept low and so does $\mathrm{pH}$.
$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O}$$
Note also the hydrogen peroxide is weakly acidic even without addition of sulphuric acid and that it's $ \mathrm{p}K_ \mathrm{a}$ depends on $\ce{H2O2}$ concentration.