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 $$f_n(x)=\tfrac1{\lambda_n}f\circ m_n(x)-f(p_n)$$ 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.
A more 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 vectors $u$. In particular parallel translation of $v(p)$ around a circle has to be close to $v(p)$. If $|v(p)|$ is large, the latter contradicts Gauss--Bonnet formula.
Comment. I want to include it in my PIGTIKAL, should I call you ccriscitiello (an anonymous mathematician)?