In the Schrödinger equation, can I have a Hamiltonian without a kinetic term?

In principle, the Hamiltonian represents the energy of a system. Whether or not you want to model your system to have kinetic energy is up to you and what you need. For example: consider an atom with an electron that can be approximated as a two level system (i.e. it as only its ground state and an excited state).

The ground state $|g\rangle$ has some energy $E_0$, and the excited state $|e\rangle$ has some energy $E_0+\Delta E$, you are always free to chose the ground state energy as you like, so chose $E_0=-\frac{\Delta E}{2}$ and you have the Hamiltonian of your system

$$H=\frac{\Delta E}{2}\left(|e\rangle\langle e|-|g\rangle\langle g|\right)=\frac{\Delta E}{2}\begin{pmatrix}-1&0\\0&1\end{pmatrix}=-\frac{\Delta E}{2}\sigma_z $$

with $\sigma_z$ the Pauli matrix. This is a perfectly valid (albeit a bit boring) Hamiltonian that has no kinetic term, basically, you decided that you don't really care about the kinetic energy of the atom, you're interested in the state of the electron. More interesting Hamiltonians that don't model kinetic degrees of freedom are given in Paradoxy's answer.

And as a note, a Hamiltonian is just a hermitian operator bounded below. There is, in principle, no further requirement. You can take any hermitian operator bounded below and start solving the Schrödinger equation. Of course, whether or not this has any physical meaning is another story.

The easiest one which comes to mind is $$H=j\sum_{<ij>} \sigma_{i}\sigma_j$$ The Hamiltonian of ising model. (They use it in Schrödinger equation too sometimes with numerical calculations)

Or as it's said in comments by @rnels12 Hamiltonian of Heisenberg model $$H=j\sum_{<ij>} \sigma_{i}.\sigma_j-h\sum_i \sigma_i$$

Edit: I am not sure why I got downvoted, but I was trying to say QM does allow this. In fact, whenever you don't care about motion of your particles or their motion are relatively small you can just drop kinetic energy term to make Hamiltonian even easier. And if you are worried about uncertainty principle or something, there won't be any problem with it if you calculate wave function and then variances. Because after all neglecting kinetic term is not equal to assuming zero momentum.

Edit 2: Thanks to @d_b I understood that the first term in hofstadter butterfly Hamiltonian is lattice version of a kinetic term, although it's not clearly $\frac{\hat{p}^2}{2m}$, in continuum limit it becomes one.

In studying the quantum Hall effect, it is common to assume that all particles have the same kinetic energy (sometimes people describe this by saying the kinetic energy is "quenched"). The assumption is reasonable when all of the particles occupy a single Landau level and the energy spacing between Landau levels (the cyclotron energy) is large relative to other energy scales in the problem. In this case, the kinetic energy just gives a constant shift in the energy spectrum, so we can ignore it and consider a hamiltonian with only an interaction term (projected to a Landau level), $H = PVP$.