I am far from being an expert, but I can confirm that there exist number fields $K\neq\mathbb{Q}$ which have no nontrivial unramified extensions. For example, imaginary quadratic number fields of class number one have this property. See this paper by Yamamura for more examples and background information.
On the other hand, I don't understand your main question. What do you mean by "reconstructing" the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$?
This is a very good question which is a big open problem. There are a number of theorems, some of them easy and some very difficult, and also a number of conjectures, restricting the class of groups which may turn out to be absolute Galois groups. But it seems that nobody has any idea about how a precise description of the class of absolute Galois groups might look like.
One general point-of-view position on which the experts seem to agree is that the right object to deal with is not just the Galois group $G_K=\operatorname{Gal}(\overline{K}|K)$ considered as an abstract profinite group, but the pair $(G_K,\chi_K)$, where $\chi_K\colon G_K\to \widehat{\mathbb Z}^*$ is the cyclotomic character of the group $G_K$ (describing its action in the group of roots of unity in $\overline K$). The question about being absolute Galois should be properly asked about pairs $(G,\chi)$, where $G$ is a profinite group and $\chi\colon G\to \widehat{\mathbb Z}^*$ is a continuous homomorphism, rather than just about the groups $G$.
For example, here is an important and difficult theorem restricting the class of absolute Galois groups: for any field $K$, the cohomology algebra $H^*(G_K,\mathbb F_2)$ is a quadratic algebra over the field $\mathbb F_2$. This is (one of the formulations of) the Milnor conjecture, proven by Rost and Voevodsky. More generally, for any field $K$ and integer $m\ge 2$, the cohomology algebra $\bigoplus_n H^n(G_K,\mu_m^{\otimes n})$ is quadratic, too,
where $\mu_m$ denotes the group of $m$-roots of unity in $\overline K$ (so $G_K$ acts in $\mu_m^{\otimes n}$ by the character $\chi_K^n$). This is the Bloch-Kato conjecture, also proven by Rost and Voevodsky.
Here is a quite elementary general theorem restricting the class of absolute Galois groups: for any field $K$ of at most countable transcendence degree over $\mathbb Q$ or $\mathbb F_p$, the group $G_K$ has a decreasing filtration $G_K\supset G_K^0\supset G_K^1\supset G_K^2\supset\dotsb$ by closed subgroups normal in $G_K$ such that $G_K=\varprojlim_n G_K/G_K^n$ and $G_K/G_K^0$ is either the trivial group or $C_2$, while $G_K^n/G_K^{n+1}$, $n\ge0$ are closed subgroups in free profinite groups. (Groups of the latter kind are called "projective profinite groups" or "profinite groups of cohomological dimension $\le1$".) One can get rid of the assumption of countability of the transcendence degree by considering filtrations indexed by well-ordered sets rather than just by the integers.
Here is a conjecture about Galois groups of arbitrary fields (called "the generalized/strengthened version of Bogomolov's freeness conjecture). For any field $K$, consider the field $L=K[\sqrt[\infty]K]$ obtained by adjoining to $K$ all the roots of all the polynomials $x^n-a$, where $n\ge2$ and $a\in K$. In particular, when the field $K$ contains all the roots of unity, the field $L$ is (the maximal purely inseparable extension of) the maximal abelian extension of $K$. Otherwise, you may want to call $L$ "the maximal radical extension of $K$". The conjecture claims that the absolute Galois group $G_L=\operatorname{Gal}(\overline{L}|L)$ is a projective profinite group.
Best Answer
Dear Tim,
As you're probably aware, this is part of the 'anabelian' etcetera.
It suffices to recover all intertia subgroups $I_v\subset H$, because their union will then be a normal subgroup $N$ such that $H/N$ is the Galois group of the maximal extension of $K$ unramified everywhere. We can get the ideal class group then by (topological) abelianization. The fact that we can get all the decomposition groups $D_v\subset G$ is Neukirch's theorem (together with Artin-Schreier at infinity). This says the maximal subgroups isomorphic to a local Galois group are exactly the decomposition groups. If you want to make this purely group-theoretic for the finite places, you invoke the theorem of Jannsen-Wingberg that lays out a presentation for all local Galois groups and consider maximal elements in the lattice of subgroups isomorphic to such an explicit presentation. Once you have the $D_v$, there is a standard group-theoretic recipe for $I_v$, which escapes me for the moment. But I'll get back to you with it, if you don't figure it out in the meanwhile.
Added:
OK, so here is the easy part. Now let $F$ be a finite extension of $\mathbb{Q}_p$ and $D=Gal(\bar{F}/F)$. We know that $D^{ab}$ fits into an exact sequence $$0\rightarrow U_F\rightarrow D^{ab}\rightarrow \hat{\mathbb{Z}}\rightarrow 0,$$ so we recover $p$ as the unique prime such that the topological $\mathbb{Z}_p$-rank of $D^{ab}$ is $r_D\geq 2$. The order $q_D$ of the residue field is 1 greater than the order of the prime-to-$p$ torsion subgroup of $D^{ab}$. Also, we know $r_D=1+[F:\mathbb{Q}_p]$. Now we apply the same reasoning to the subgroups of finite index in $D$ to figure out those corresponding to unramified extensions. That is, consider the subgroups $E$ of finite index such that $q^{r_D-1}_E=q_D^{r_E-1}$. Then the inertia subgroup of $D$ is the intersection of all of these.