Firstly, let us assume that we are restricted to measurements in the computational basis, i.e. outcomes of $|0\rangle$ or $|1\rangle$.
Next, since neither gates create any superposition (when applied to computational basis states) we can observe that without a Hadamard gate, our qubit state is always in a single computational basis state.
Hence, we can see that the computational power of our CNOT + phase gate set is not greater than that of just a simple CNOT, and so we can reduce the initial problem to simply considering the computational power of the classical CNOT.
So, in answer to the first question, a quick bit of googling reveals the answer is no, since the CNOT gate is not universal for classical computation, reversible or otherwise.
In answer to your second question, I guess the best answer I can give is simply that it is the same as the set produceable by circuits of CNOTs. I am not sure if this set of Boolean functions is well characterised or not, but perhaps this is worth posing in the CompSci stack exchange.