$\mathsf{PA}$ is vastly more than enough to prove that the halting problem is incomputable. $\mathsf{PA}$ is in fact massive overkill for the vast majority of theorems about natural numbers, or "finitary objects" (such as Turing machines, despite the infinite tape present in our intuitive conception of them) more generally.
Meanwhile, in some sense $\mathsf{Q}$ is too weak a system for the question to be meaningful. There is very little which $\mathsf{Q}$ can prove at all - e.g. it cannot prove that every number is either even or odd (= $1$ less than an even number), nor can it prove that addition is commutative. Among other things this means that intuitively-trivial modifications to definitions often wind up resulting in inequivalent versions "over $\mathsf{Q}$." For example, I don't believe that $\mathsf{Q}$ proves that $2$-tape Turing machines are no stronger than $1$-tape machines. At this point it's not clear that the question means what we want it to.
Here's a sketch of why the incomputability of the halting problem goes through in a very weak system of arithmetic. (There are various versions of the halting problem; I'll use $\{e:\Phi_e(0)\downarrow\}$, where $\Phi_e$ is the $e$th Turing machine in the standard enumeration and "$\downarrow$" means "halts," for simplicity.)
The proof boils down to basic logic + a machine construction: given a machine $\Psi$ we need to define an auxiliary machine $\hat{\Psi}$ to "defeat" $\Psi$ as a putative halting problem solver. This $\hat{\Psi}$ behaves as follows: on input $e$ it simulates $\Psi(e)$; once (if ever) this simulated computation halts, $\hat{\Psi}$ goes into an endless loop if the result is $1$ and otherwise halts and outputs $0$ (say).
So all we need is a theory of arithmetic capable of proving that the construction $\Psi\leadsto\hat{\Psi}$ actually "makes sense." Since it's an explicit construction this boils down to a totality claim, and at a glance knowing that polynomial functions are total should already be enough. So we get all the way down to theories of bounded arithmetic, which form incredibly tiny fragments of $\mathsf{PA}$.
The halting problem asks whether there exists a (single) Turing machine $H$ such that for every Turing machine $T$ and input $I$, when we feed $\langle T,I\rangle$ as input to $H$ then
- $H$ will terminate with an answer Yes or No
- the answer will reflect whether $T$ halts upon $I$
You suggest that instead we ask for a $H_T$ for every $T$? Or even go further and ask for a $H_{\langle T,I\rangle}$ for every pair $\langle T,I\rangle$? The latter would be easy! One of
PRINT "YES"
or
PRINT "NO"
will do the job, but I won't tell you beforehand which of these!
So back to the other idea: Suppose that for every TM $T$, there exists a TM $H_T$ such that $H_T$ halts for every input $I$ and gives us either a "YES" or "NO" that correctly tells us whether $T$ will halt on input $I$. Then in particular, there exists $H_S$ for the simulation TM $S$ (which we can write down explicitly), i.e., $S$ takes the description of a TM $T$ as input and simulates what TM $T$ would do when given $T$ as its input; in particular, $S$ halts on input $T$ iff $T$ halts on input $T$.
Now from $H_S$ we can explicitly construct yet another TM $Z$, which is just $H_S$ except that every terminating node is replaced with "check what output we produced; if it is 'NO', terminate; otherwise, loop forever".
What might $Z(Z)$ produce? That depends on $H_S(Z)$, which again depends on $S(Z)$, so ultimately on $Z(Z)$! If $Z$ halts on input $Z$, then $S$ halts on input $Z$, then $H_S$ halts on input $Z$ with result "YES", then $Z$ does not halt on input $Z$ - contradiction. On the other hand, if $Z$ does not halt on input $Z$, then $S$ does not halt on input $Z$, then $H_S$ halts on input $Z$ with answer "NO", then $Z$ halts on input $Z$ - contradiction again!
We conclude that $H_S$ cannot exist.
Hence there exists at least one TM that does not have its specific halt-checker TM. Hence it is false that every TM has a halt-checker TM.
As to the undecidability of math: For any given statement $\phi$, we (i.e., a TM) can enumerate all finite sequences of math symbols and check whether such a sequence constitutes a proof of either $\phi$ or a proof of $\neg \phi$. Such a proof-finding TM, let's call it $P$, will either halt with a proof for $\phi$ or with a proof for $\neg \phi$ - or not halt at all. But perhaps we are lucky and this will always halt? Unfortunately, for every given TM $T$ and input $I$, we can formulate "TM $T$ will halt on input $I$" as a mathematical statement. Thus $P$ above will either find a proof that $T$ halts or a proof that it does not halt on input $I$. Per halting problem, we know that this is impossible. Therefore, there do exist some inputs for which $P$ does not halt. That is, there do exist some statements $\phi$ such that neither $\phi$ nor $\neg\phi$ is provable.
Best Answer
Even assuming we're okay drastically restricting the scope of mathematics to be decided - which flies in the face of the whole idea of the program - there's still a fundamnetal issue: the arithmetizability of computation. Basically, in order to rescue decidability you'd have to restrict all the way down to triviality.
This is a riff on Godel's incompleteness theorem. Just as with GIT we used arithmetic to talk about proofs, we can talk about Turing machines in the language of arithmetic. (I don't know when this was first explicitly stated, but it's pretty much immediate once Godel's ideas are understood - certainly Kleene's $T$-predicate, which appeared only seven years after Turing's paper, does the job.) The point is that from a computable solution to the Entscheidungsproblem we can construct a computable consistent completion of PA (just go through sentences one by one with a consistency-preserving greedy algorithm), but by the arithmetization of computation (plus Rosser's improvement of GIT as originally stated) Godel's incompleteness theorem applies to any computably axiomatizable consistent extension of PA.
Indeed, looking a bit closer we get that even the $\Sigma_1$ fragment of the consequences of PA is not computable. Since the $\Delta_0$ fragment (= bounded quantifiers only) is trivially computable, this says that incompleteness hits arithmetic as hard as it possibly could. And at this point there's really nothing left to do.
Combining all this with the fact that restricting the scope goes against the whole spirit of the program to begin with, I think that explains why the conclusion was (and is) pretty universal. Note that the study of decidable fragments of arithmetic is definitely alive and well, it's just recognized as a fundamentally different thing than Hilbert's program.