The function $f$ has to be Lipschitz.
Assume contrary, choose a sequence of points $p_n$ such that $\lambda_n=|\nabla_{p_n}f|\to \infty$. Let $f_n=\tfrac1{\lambda_n}f\circ m_n$, where $m_n$ is a motion of the plane that sends $p_n\mapsto p$.
The Hession of $f_n$ converges to zero. So, after passing to a subsequence we can assume that $f_n$ converges to an affine function, say $f$ (it is convex and concave at the same time). Note that $|\nabla_p f|=1$. So, there is no such function on the hyperbolic plane --- a contradiction.
Postscript. I just realized that this question was already asked by the same OP and I already gave an answer before. Surprisingly the answer is nearly identical.