Neutralization between Ca(OH)2 + H2SO4 in a 30% hydrogen peroxide solution
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.

$$\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$

where $\mathrm{p}K_  \mathrm{a,\ce{H2O2}}=11.75$$ by Wikipedia, but see the links below.

If we consider reaction

$$\ce{Ca(OH)2 +  H2O2 -&gt;  H2O + Ca(OH)(HO2)}$$

we need to neutralize 50% of $\ce{H2O2}$ to reach $\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}}$

The hydroxide forms from $\ce{H2O2}$  the $\mathrm{pH}$ buffer solution of a weak acid and it's salt.

$$\begin{align}
\ce{Ca(OH)2  &amp;&lt;=&gt;&gt; CaOH+  + OH-}\\
\ce{CaOH+ &amp;&lt;=&gt;&gt; Ca^2+  + OH- }\\
\ce{H2O2 &amp;&lt;&lt;=&gt; H+ + HO2-}\\
\ce{H+ + OH- &amp;&lt;=&gt;&gt; H2O}\\
\end{align}$$

$\ce{Ca(OH)2}$: $\mathrm{p}K_\mathrm{b1} =1.37$, $\mathrm{p}K_\mathrm{b2} =2.43$
( Wikipedia )

Additionally, $\ce{HO2-}$ is partially eliminated by precipitation, therefore ratio $ \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$ is kept low and so does $\mathrm{pH}$.

$$\ce{CaOH+ + HO2- + 7 H2O &lt;=&gt;&gt; CaO2 \cdot 8 H2O v}$$

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.

H2O2 pH-and-Ionization-Constant

The solubility constant of calcium peroxide octahydrate in relation to temperature; its influence on radiolysis in cement-based materials