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.)

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.)