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 $$f_n(x)=\tfrac1{\lambda_n}\cdot f\circ m_n(x)-f(p_n)$$ where $m_n$ is a motion of the plane that sends $p\mapsto p_n$.
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.
Мore direct proof. Suppose $f$ has bounded Hessian. Observe that the gradient $v=\nabla f$ is almost parallel; that is, there is a constant $C$ such that $|\nabla_u v|\le C$ for any unit tangent vector $u$. In particular, the parallel translation of $v(p)$ around a circle has to be close to $v(p)$. If $|v(p)|$ is large, the latter contradicts the Gauss--Bonnet formula.
Comment. It is a nice problem --- I plan to include it in my PIGTIKAL.