Disclaimer
This answer is somewhat an addition to the existing ones. Here, I will try to present a practically feasible method for obtaining lead(IV) oxide $\ce{PbO2}$ from the available reagents. It is a dark brown precipitate, almost black, as required in the question.
Required reagents
The following reagents from the list are needed in order to get the desired black precipitate of lead dioxide: lead(II) nitrate $\ce{Pb(NO3)2}$, hydrogen peroxide $\ce{H2O2}$ (the higher the concentration, the better) and sodium hydroxide $\ce{NaOH}$. All solids have to be dissolved in water while sodium hydroxide solution should not be too strong.
Procedure
At first, it is necessary to mix the solutions of sodium hydroxide and hydrogen peroxide, obtaining an alkaline $\ce{H2O2}$ solution. Then, when the alkaline solution of hydrogen peroxide is added to a $\ce{Pb(NO3)2}$ solution, a precipitation of dark brown solid occurs according to the following reaction equation:
$$\ce{Pb(NO3)2 (aq) + H2O2 (aq) -> PbO2 (s) + 2HNO3 (aq)}$$
Heating can help to speed up the reaction, but too much heating can dissolve the precipitate:
$$\ce{PbO2 (s) + 2OH- (aq) + 2H2O (l) \xrightarrow{\textrm{heating}} [Pb(OH)6]^2- (aq)}$$
Reference
- Ronald L. Rich. Inorganic Reactions in Water (1st ed.): "Peroxide", p. 356.