For which Millennium Problems does undecidable -> true?

$P\ne NP$ is a $\Pi^0_2$ statement:

for each polynomial $p$ and Turing machine $M$ implementing an algorithm attempting to decide SAT, there is a formula $\phi_M$ such that if we look at the computation of $M$ on input $\phi_M$ after $p(|\phi_M|)$ computation steps, $M$ has either not halted or it has answered incorrectly.

(Edit: added more detail to the $\Pi^0_2$ statement above.)

If we place a computable bound on the size of $\phi_M$ as a function of $M$ then we can improve this to a $\Pi^0_1$ statement, and hence will have the property $\text{undecidable}\rightarrow\text{true}$, since true $\Sigma^0_1$ statements are provable (in Peano Arithmetic). Of course in a formal sense, any provable statement is equivalent to $0=0$ which is $\Pi^0_1$.

Is it plausible that that $P\ne NP$, but there is no computable bound on $|\phi_M|$? This would mean that there are algorithms that do "arbitrarily well" at solving SAT in polynomial time. That is, relative to their description, they are correct for an Ackermann-function or busy-beaver-function level many $\phi$'s.

But intuitively, there is not much structure in SAT to exploit, and so once you have so many variables $x_1,\ldots,x_v$ that a random assignment of truth values has higher Kolmogorov complexity than $M$: $$K(M)\le v + O(1) $$ then there ought to be a $\phi(x_1,\ldots,x_m)$ on which $M$ fails. Since $K(M)$ has a computable bound as a function of $M$, it would follow that $M\mapsto |\phi_M|$ is computably bounded. This is probably off by a "factor" somewhere; maybe a better way is to say that $P^A\ne NP^A$ for a random oracle $A$, and in that case $\phi_M$ is computably bounded.

So we can make a plausible "computably bounded $P\ne NP$ conjecture" that $M\mapsto \phi_M$ is computably bounded (with a specific bound being incorporated into the conjecture), which does have the property $\text{undecidable}\rightarrow\text{true}$.

Global regularity for Navier-Stokes on the torus is (logically equivalent to) a $\Pi_2^0$ statement; this is essentially an unpublished observation of Bourgain (who made it in the more general context of supercritical equations), which I sketched out in this paper http://arxiv.org/abs/0710.1604 . Basically, Navier-Stokes is equivalent to the assertion that for all $T, E > 0$, there exists an $M$ such that all initial data with $H^1$ norm at most $E$, there exists a solution up to time $T$ whose $H^1$ norm is always bounded by $M$. Now if such a claim is the case, it can be verified (using rigorous perturbation theory) by constructing a sufficient number of approximate solutions to a sufficient number of choices of initial data, and verifying that all of these approximate solutions have norm at most $M/2$ (say) up to time $T$, where "sufficient number" is something explicit that depends on $T, E, M$, and the nature of the approximation can be discretised at an explicit scale that also depends on $T,E,M$. So Navier-Stokes is equivalent to a statement of the form $\forall T,E \exists M: P(T,E,M)$ where $P(T,E,M)$ is something that can be verified in finite time for each $T,E,M$ (which can be taken to be rational numbers). (In fact, for each $E$ there is an explicit time $T = T(E)$ for which one needs to verify $P(T,E,M)$, and beyond which global regularity is easily obtained from the energy dissipation, so the claim even simplifies a little to $\forall E \exists M: P( T(E), E, M)$.)

It looks very difficult to me, however, to make Navier-Stokes equivalent to a $\Pi^0_1$ statement. This would basically amount to being able to describe all possible "blowup scenarios" by a countable set, and to be able to determine whether each such blowup scenario can actually happen in finite time. While some blowup scenarios (particularly "stable", "approximately self-similar" ones) can be described and verified in such a manner, it's not clear whether all of them can.

As Harry Altman pointed out, for a conjecture undecidable -> true means that it can be formulated as a $\Pi_1^0$ statement. To put it simply, if the conjecture is false, one can prove this by an explicit (finite) calculation. I would leave Yang–Mills and Navier–Stokes to someone more familiar with mathematical physics then I am. The other three conjectures, apparently, aren't $\Pi_1^0$ statements for now.

By the way, the Poincare conjecture wasn't a $\Pi_1^0$ statement before Rubinstein's algorithm was discovered (which determines whether or not a 3-manifold is the 3-sphere, see the comment by HJRW). (Expert guys, fix me here if I am wrong). So, whether or not a particular conjecture is in this calss depends on the awailable knowledge. After all, once a conjecture is proven, it is in $\Pi_1^0$ by definition.

Let me begin with Birch and Swinnerton-Dyer Conjecture. Technically, what you need to get \$1M is to prove or disprove the following (http://www.claymath.org/millenium-problems/birch-and-swinnerton-dyer-conjecture). If $E$ is an elliptic curve over $\mathbb{Q}$, then $r = rank(E(Q))$, called arithmetic rank, is equal to the order $r_*$ of zero of $L(E, s)$ at $s=1$, called analytic rank. This conjecture actually consists of two rather different parts, namely $r\le r_*$ and $r\ge r_*$. The first one is a $\Pi_1^0$ statement. If $r>r_*$ for a particular curve $E$, you can prove it by a direct computation (with some luck). The other half of the conjecture is more tricky, as Will Sawin pointed out. The problem is, there is no known algorithm to compute the group $E(Q)$. (Do not take me wrong: there are some algorithms, and they apparently work. We just don't have a proof that they work.)
Theoretically, it is possible that while $r<r_*$ for some curve $E$, you will never know it because you won't be sure if there are some more generators of $E(Q)$ which you did not find yet. In fact, Manin used this very argument in the opposite direction, and proposed an algorithm of computing the Mordell-Weil group assuming the Birch and Swinnerton-Dyer Conjecture to be true. (See Hindry, Silverman "Diophantine Geometry" for details. To be pedantic, you need a bit more then $r=r_*$ for this.)
So, the inequality $r\ge r_*$ is not a $\Pi_1^0$ statement yet, for the best of my knowledge.

I can't say anything sensible about the Hodge conjecture, except that it definitely does not look like $\Pi_1^0$. In order to disprove it by an explicit example, you need to prove that on a particular variety there is no algebraic cycles with a given cohomology class. Maybe experts know better, but I have never heard about any algorithm for a job like this.

Bjørn Kjos-Hanssen already gave an answer about $P\neq NP$, and I don't have much to add. Except for, as I pointed out in a separate question, How to prove a $\Pi_2$ statement properly?, there is a technical possibility that that problem is decidable, but the "decision" is wrong. (Do not take me seriously, I don't really believe in this.)