The answer is quite simple: When you have $\ce{H2O}$, it means you have two hydrogen and one oxygen atoms. Each hydrogen brings 1 electron, while the oxygen brings 6.
Your assumption now is, that you have another hydrogen atom bringing in another electron. In this case, you'd be right to assume that there are three electrons from hydrogen, totaling up to 9 electrons in the molecule. But this would violate the octet rule and consequently Paulis exclusion principle.
The solution to your question is, that you do not have $\ce{H3O}$, but $\ce{H3O+}$! The plus means, there is one electron less in the molecule, because you don't bring another H (with one electron) into the molecule, but rather just a proton ($\ce{H+}$ with the electron removed).
So, you have 6 electrons from the oxygen atom and only two from the hydrogen atoms, totaling up to 8 electrons. Three covalent bonds take up 2 electrons each, leaving you with two electrons in the remaining sp3 hybrid orbital as the lone pair.