We can easily construct such a function $f$, and indeed, there are continuum many such functions. The reason is that you've imposed only countably many requirements on $f$, which can each be satisfied by a finite piece of $f$, one after the other. So we can satisfy those requirements one-by-one.

Let's do the "forcing" manner of construction, which has an affinity with the Baire category theorem. That is, consider the space of all finite partial functions $p:n\to\mathbb{N}$, essentially the finite sequences of natural numbers, ordered by extension. For each computable functions $g$ and $h$, where $g$ is increasing, let $D_{g,h}$ be the set of such finite functions $p$ for which $p(g(n))\neq h(n)$ for some $n\in\text{dom}(p)$, that is, the set of conditions $p$ that have already satisfied the requirement imposed by $g$ and $h$. Any extension of such a $p$ to a total function $f$ will also satisfy the requirement impose by $g$ and $h$. The key thing to notice is that any condition $p$ can easily be extended to a condition $p'\in D_{g,h}$, because there will be some $n$ such that $p(g(n))$ is not yet defined, and we can extend $p$ to some $p'$ by defining $p'(g(n))$ in such a way that makes it different than $h(n)$, so that $p'\in D_{g,h}$. Another way to say this is that the sets $D_{g,h}$ are dense (and open) in the space of all finite partial functions. So we have countably many open dense requirements to meet. We can simply meet them one by one. That is, we can build a sequence of finite extensions $$f_0\subset f_1\subset f_2\cdots$$
such that the union function $f=\bigcup_n f_n$ has initial segments meeting all the requirements $D_{g,h}$. It follows that for every $g$ and $h$ there is some $n$ for which $f(g(n))\neq h(n)$ and so $f$ is absolutely random.

If you think carefully about it, what we have done in effect is to build a tree of finite approximations, such that every branch through this tree will be absolutely random. Thus, the construction provides continuum many absolutely random functions $f$.