Claim 1: The first half of what I said in my comments is correct: for each subset $S$ of the prime numbers, there is a field $F$ having the property that it admits a degree $d$ field extension iff $d$ is divisible only by primes in $S$.

As Qiaochu says, this comes about because of the existence of perfect fields $K$ with absolute Galois group $\hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$. In particular, take the closed subgroup $H_S = \prod_{p \not \in S} \mathbb{Z}_p$, and let $K_S = \overline{K}^{H_S}$.
We can take $K$ to be any finite field, or $\mathbb{C}((t))$, for instance.

The part of my claim about using $\mathbb{R}((t))$ to get exact $2$-divisibility seems not necessarily correct to me now. The problem is that the absolute Galois group of $\mathbb{R}((t))$ is a nonabelian extension of $\hat{\mathbb{Z}}$ by $\mathbb{Z}/2\mathbb{Z}$: indeed it is the profinite completion of the infinite dihedral group.

As for what I said about the converse: no way, things are definitely more complicated than that! What I had in mind was the "local" observation that by Artin-Schreier, for any prime $p$ the Sylow-$p$-subgroup of the absolute Galois group is either trivial, infinite or has order $2$. (The case of characteristic $p > 0$ has to be given a little separate attention, but I don't think there's a problem there.)

However, the different Sylow p-subgroups will not act independently unless the absolute Galois group is pro-nilpotent. (Note that the profinite dihedral group is not pro-nilpotent.) Thus:

Claim 2: If $K$ has characteristic $0$ and pro-nilpotent Galois group, then the possible orders are exactly those in Claim 1.

Here is an explicit example to show that the converse to Claim 1 is not generally correct: let $F$ be the maximal solvable extension of $\mathbb{Q}$. Then it has no quadratic extensions. However, it certainly does have extensions of even degree, since otherwise -- e.g. by Feit-Thompson! -- the absolute Galois group of $\mathbb{Q}$ would be pro-solvable, which it most certainly is not: many nonabelian simple groups (e.g. $A_n$ for $n \geq 5$) are known to occur as Galois groups over $\mathbb{Q}$.

What I said is the correct answer (with 2 inverted!) to a different question: which supernatural numbers can be the order of the absolute Galois group of a field?

Claim 1: The first half of what I said in my comments is correct: for each subset $S$ of the prime numbers, there is a field $F$ having the property that it admits a degree $d$ field extension iff $d$ is divisible only by primes in $S$.

As Qiaochu says, this comes about because of the existence of perfect fields $K$ with absolute Galois group $\hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$. In particular, take the closed subgroup $H_S = \prod_{p \not \in S} \mathbb{Z}_p$, and let $K_S = \overline{K}^{H_S}$.
We can take $K$ to be any finite field, or $\mathbb{C}((t))$, for instance.

The part of my claim about using $\mathbb{R}((t))$ to get exact $2$-divisibility seems not necessarily correct to me now. The problem is that the absolute Galois group of $\mathbb{R}((t))$ is a nonabelian extension of $\mathbb{Z}/2\mathbb{Z}$ by $\hat{\mathbb{Z}}$: indeed it is the profinite completion of the infinite dihedral group.

As for what I said about the converse: no way, things are definitely more complicated than that! What I had in mind was the "local" observation that by Artin-Schreier, for any prime $p$ the Sylow-$p$-subgroup of the absolute Galois group is either trivial, infinite or has order $2$. (The case of characteristic $p > 0$ has to be given a little separate attention, but I don't think there's a problem there.)

However, the different Sylow p-subgroups will not act independently unless the absolute Galois group is pro-nilpotent. (Note that the profinite dihedral group is not pro-nilpotent.) Thus:

Claim 2: If $K$ has characteristic $0$ and pro-nilpotent Galois group, then the possible orders are exactly those in Claim 1.

Here is an explicit example to show that the converse to Claim 1 is not generally correct: let $F$ be the maximal solvable extension of $\mathbb{Q}$. Then it has no quadratic extensions. However, it certainly does have extensions of even degree, since otherwise -- e.g. by Feit-Thompson! -- the absolute Galois group of $\mathbb{Q}$ would be pro-solvable, which it most certainly is not: many nonabelian simple groups (e.g. $A_n$ for $n \geq 5$) are known to occur as Galois groups over $\mathbb{Q}$.

What I said is the correct answer (with 2 inverted!) to a different question: which supernatural numbers can be the order of the absolute Galois group of a field?