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.

$$\mathrm{pH}=\mathrm{p}K_  \mathrm{a} + \log \frac{[\ce{A-}]}{[\ce{HA}]}$$

As there is the weak acid excess over the hydroxide, the hydroxide is used to form  the buffer solution of hydrogen peroxide and it's salt.

$$\ce{H2O2 <<=> H+ + HO2-}$$
$$\ce{H+ + OH- <=>> H2O}$$


Additionally, if the conjugate basis $\ce{A-}$ is eliminated by precipitation, the ratio $ \frac{[\ce{A-}]}{[\ce{HA}]}$ is kept low and so does $\mathrm{pH}$.

$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O}$$

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.

[H2O2 pH-and-Ionization-Constant][1]

[The solubility constant of calcium peroxide octahydrate in relation to temperature][2]


  [1]: http://www.h2o2.com/technical-library/physical-chemical-properties/thermodynamic-properties/default.aspx?pid=51&name=pH-and-Ionization-Constant
  [2]: https://www.sciencedirect.com/science/article/pii/S0022311511008440