There are some nice examples in knot theory and quantum algebra.
If $X$ is an algebraic variety over $\mathbb{Q}$, there is a canonical outer action $G_{\mathbb{Q}} \rightarrow Out(\hat{\pi}_1(X(\mathbb{C})))$ (which also exists for other fields than $\mathbb{Q}$ using the so-called algebraic fundamental group but I don't want to say something wrong about that). Roughly speaking, this is because finite covering of $X$ can be defined over $\bar{\mathbb{Q}}$, together with the relation between (regular) finite covering and finite quotients of the fundamental group. This action thus has the same origine as for dessin d'enfants.
Of particular interest is the study of this action in the case $X$ is the moduli space of algebraic curves of genus $g$ with $n$ marked points. This was suggested in Grothendieck's esquisse, and leads to the so-called Grothendieck-Teichmuller theory which gives a rather explicit description of a group which actually contains $G_{\mathbb{Q}}$. (see Cartographic group and flat stringy connection or Where is a good place to start learning about the Grothendieck-Teichmuller group?)
Actually, there are several flavours of the Grothendieck-Teichmuller group: a profinite one $\widehat{GT}$ which does contain $G_{\mathbb{Q}}$ and a group $GT(k)$ defined for every field $k$. A deep result of Drinfeld assert that this latter group is in some sense a universal automorphism group of braided monoidal categories. Indeed, it acts on the set of Drinfeld associator with coefficients in $k$.
Now, there is also a morphism
$G_{\mathbb{Q}}\rightarrow GT(\mathbb{Q}_{\ell})$
for every prime number $\ell$. Hence the absolute galois group acts on each kind of object in which associators come up (assuming that it is in a situation where one can work over $\mathbb{Q}_{\ell}$). It leads to, I think, quite surprizing examples like action on finite type invariants of knots and links, on quantization functor of Lie bialgebras, and several other constructions arising in deformation/quantization theory as they are often related to Drinfeld associators.
[Edit:] The answer should be positive, that is, every profinite group appears as the fundamental group of a scheme. Here is a sketch of proof.
First of all, I claim that for any finite group $G$ there exists a complex affine simply connected variety $X$ with a free action of $G$. Start from a faithful finite-dimensional representation $G \to \mathrm{GL}(V)$ of $G$, such that there is an open subset $U \subseteq V$ where the action of $G$ is free, and such that $V \smallsetminus U$ has codimension at least 2 in $V$. Then $X$ is obtained with an easy equivariant extension of Jouanolou's trick. This yields a $G$-covering $X \to X/G$; then $X/G$ is an affine variety, and its fundamental group is $G$.
Now take a profinite group $G = \projlim_{i\in I}G_i$; identify $G$ with the corresponding affine group scheme over $\mathbb C$, in the usual fashion. For each finite subset $J\subseteq I$ denote by $G_J$ the image of $G$ in $\prod_{j\in J}G_j$; clearly we have $G = \projlim_{J \subseteq I}G_J$.
For each $i$ take a complex affine simply connected variety $X_i$ with a free action of $G_i$. Consider the affine scheme $X := \prod_{i \in I}X_i$, with the resulting action of $G$, and the quotient $X/G$, which is the spectrum of the ring of invariants $\mathbb C[X]^G$. For each finite subset $J\subseteq I$ set $X^J := \prod_{j \in J}X_j$. The action of $G$ on $X^J$ factors through a free action of $G_J$. Furthermore, we have $\mathbb C[X]^G = \injlim_{J \subseteq I}\mathbb C[X^J]^{G_J}$, hence $X/G = \projlim X^J/G_J$.
Now, $X^J$ is simply connected, hence the fundamental group of $X^J/G_J$ is $G_J$. On the other hand, it is easy to see that the Galois category of $X$ is the inductive limit of the Galois categories of the $X^J$, so that its Galois group is precisely $\projlim G_J = G$.
[Edit2:] Let me clarify what I mean by the "equivariant Jouanolou trick". The theorem of Jouanolou says that if $U$ is a quasi-projective variety, there exists an locally trivial fibration in affine spaces $X\to U$, where $X$ is affine. What we need here is the statement that if $G$ is a finite group acting on $U$, then we can construct such a map $X \to U$ that is also $G$-equivariant. This is easy: start from a fibration in affine space $Y \to U$ with $Y$ affine, and take $X$ to be the fiber product over $U$ of all the $g^*Y$ for $g \in G$, with the obvious action of $G$.
Best Answer
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.