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.** I want to include it in my [PIGTIKAL][1], should I call you *an anonymous mathematician ccriscitiello*?


  [1]: https://arxiv.org/abs/0906.0290