in your question formulation, you have forgotten to take into account $\ce{H2O2}$ is a weak acid.
The title should rather be: Neutralisation between calcium hydroxide and 30% hydrogen peroxide"
Unless $\ce{Ca(OH)2}$ was in excess over $\ce{H2O2}$ - and it was said it was not - $\mathrm{pH}$ would be always significantly lower than pH of the hydroxide.
$$\mathrm{pH}=\mathrm{p}K_ \mathrm{a,\ce{H2O2}} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$
The hydroxide forms from $\ce{H2O2}$ the $\mathrm{pH}$ buffer solution of a weak acid and it's salt.
$$\begin{align} \ce{Ca(OH)2 &<=>> CaOH+ + OH-}\\ \ce{CaOH+ &<=>> Ca^2+ + OH- }\\ \ce{H2O2 &<<=> H+ + HO2-}\\ \ce{H+ + OH- &<=>> H2O}\\ \end{align}$$
$\ce{Ca(OH)2}$: $\mathrm{p}K_\mathrm{b1} =1.37$, $\mathrm{p}K_\mathrm{b2} =2.43$ ( Wikipedia )
Additionally, $\ce{HO2-}$ is partially eliminated by precipitation, therefore ratio $ \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$ is kept low and so does $\mathrm{pH}$.
$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O v}$$
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.