Edit: There's an additional subtlety I feel like I've glossed over which I'd like to describe in more detail. First, we can write down a predicate $\text{Computes}(T, r)$ (in first-order ZF, to be specific) which given a Turing machine $T$ and a real number $r$ asserts that $T$ computes $r$.
Now, "$r$ is computable" is defined to mean "$\exists T : \text{Computes}(T, r)$." So an important part of the potential confusion here has to do with the standard meaning of $\exists$ in classical first-order logic: importantly, proofs of existence are not required to be constructive, in the sense that a proof that $\exists x P(x)$ is not required to proceed by constructing an explicit element $x$ and then verifying that it satisfies $P(x)$. So it is possible to prove nonconstructively that a real number, such as $BB(6)$, is computable, without constructing an explicit Turing machine computing it.
You might consider this unsatisfying and a bad way to define "computable," or at least you might want to be able to refer to some notion of "explicit computability." (If so you might be interested in reading about constructivism.) Here's a simple way to do it: we just define an explicit computation of a real number $r$ to be a construction of a Turing machine $T$ satisfying $\text{Computes}(T, r)$ (one must supply both the Turing machine $T$ and a proof that it computes $r$). In other words, to prove that a real number $r$ is "explicitly computable" one must exhibit the Turing machine computing it, rather than only proving that some such Turing machine exists.
Note, however, that I really do want to alter the grammar here to "explicit computation" rather than "explicitly computable." This is not a definition of a class of real numbers satisfying a logical condition; this notion of "explicit computability" depends on which Turing machines have been written down so far, and in particular it changes over time if some new Turing machine is written down which explicitly computes some number that had not yet been previously computed. Also, only finitely many real numbers have been explicitly computed in this sense since only finitely many Turing machines have been written down. Worse, only finitely many real numbers can be explicitly computed in this sense, since the universe (as far as we know) is not arbitrarily large and cannot fit descriptions of arbitrarily large Turing machines in it.
However, this is a way to make precise the sense in which we don't know $BB(6)$: nobody has yet written down a Turing machine computing it, so it has not been explicitly computed in the sense above.
(Original answer follows from here.)
All the online resources that I've seen on uncomputable numbers assume that they're all irrational.
This isn't an assumption, it's a theorem. Every rational number is computable because there exists a program which prints out the digits of any rational number. Similarly, because we can compute the roots of a polynomial with rational coefficients to arbitrary precision, every algebraic number is computable, so uncomputable numbers are transcendental.
But doesn't this also describe the outputs of uncomputable functions? Busy_beaver(6) cannot be computed by any finite terminating algorithm, yet it's an integer.
We need to distinguish carefully here between objects and descriptions of objects. We have no idea what number $BB(6)$ is, but if you believe that $BB$ is a well-defined function, whatever number $BB(6)$ is, that number is trivially computable, namely by a program which prints out its digits. We simply don't know which program this is!
This issue actually has nothing to do with $BB$ not being computable, it is really about the difference between numbers and descriptions of numbers. For example, consider the integer which is equal to $0$ if the Riemann hypothesis is false and $1$ if the Riemann hypothesis is true. Whichever number this is, it is again trivially computable, by either the program which prints $0$ or the program which prints $1$. We just don't know which program is correct!
(Less trivially, there is a single computer program, which in principle we can write down explicitly right now, which prints $0$ if the Riemann hypothesis is false and prints $1$ if it's provable in ZFC. This is because we can search through the nontrivial zeroes of the zeta function and verify that they have real part $\frac{1}{2}$ (it requires some nontrivial complex analysis to show that this is possible), and we can also search through proofs of the Riemann hypothesis in ZFC. However, this program would run forever in the unlikely event that the Riemann hypothesis is true but independent of ZFC.)
For that matter, how do we know that Chaitin's Constant isn't the result of any finite terminating function? The argument given on the Wikipedia page only proves that we can never compute what number is Chaitin's Constant, but it doesn't prove that Chaitin's Constant is itself not the result of any other unrelated function. (For example, how can you prove that Chaitin's Constant isn't pi/4? Maybe it is, we'd just never know.)
No, the argument on Wikipedia proves that the digits of Chaitin's constant cannot be printed by any Turing machine (this is what it means for a specific number to not be computable), because if it could then that Turing machine also solves the halting problem. Since the digits of $\frac{\pi}{4}$ (or whatever other computable real you like) can be printed by a Turing machine, it cannot be equal to Chaitin's constant. This is a genuinely different situation from the $BB(6)$ example, which can be printed by an unknown Turing machine.
The following analogy may be helpful: suppose somehow that someone proved by contradiction that $\pi + e$ is algebraic (this number is not known to be transcendental). This would mean that there is some rational polynomial $f(x)$ such that $f(\pi + e) = 0$. This would be true even if the proof does not produce $f$, so we don't know what it is. So this unknown polynomial would be analogous to the unknown Turing machines above.
