Main question: Suppose we have some real number $x$ that is proven to be not only be computable, but has had such an algorithm explicitly found. Could it be the case that it may be impossible to prove or disprove the rationality of $x$?
Along the same notion, if we have proven that a number's rationality status is knowable, does that prove the number is computable?
More generally, how is the knowledge of computability of a number related to the ability of knowing its rationality status?
This is a strange thought, and I have absolutely no intuition towards or against such a possibility.
In order to talk about provability/decidability we need to specify a background set of axioms - no statement is undecidable in all systems, for silly reasons. I'm going to assume we're working in $\mathsf{ZFC}$, but what follows isn't specific to $\mathsf{ZFC}$ at all.
For a sentence $\varphi$, define a sequence of natural numbers $k_n$ ($k\in\mathbb{N}$) as follows: $k_n=1$ iff EITHER $n$ is a power of $2$ OR $\varphi$ is provable from $\mathsf{ZFC}$ in $\le n$ steps, and $k_n=0$ otherwise. The number $$\xi_\varphi:=\sum_{n\in\mathbb{N}}{k_n\over 2^n}$$ is then guaranteed - provably in $\mathsf{ZFC}$, or indeed much less - to be rational iff $\varphi$ is provable from $\mathsf{ZFC}$.
Now a quick corollary to Godel's theorem shows that the general problem of deciding whether a sentence is $\mathsf{ZFC}$-provable is undecidable (basically, otherwise we could use a greedy algorithm to whip up a consistent computable completion of $\mathsf{ZFC}$). So there is in general no way to decide whether $\xi_\varphi$ is rational. In fact this can be made explicit: for each "reasonable" theory $T$ we can whip up a specific sentence $\psi_T$ such that $T$ cannot decide whether $\xi_{\psi_T}$ is rational or not.
(Note that the above has essentially nothing to do with rationality per se - pretty much any nontrivial property of numbers will be subject to some variant of this argument.)