Is there a "unique" homogeneous contact structure on odd-dimensional spheres?
Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.
Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.
By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.
By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n{+}1),\quad \mathrm{SU}(n{+}1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ (thought of as column vectors of height $\tfrac12(n{+}1)$ with quaternion entries) and uses the same formula as before, but now, one allows $J$ to be scalar multiplication (on the right) by any unit imaginary quaternion. Upon conjugating by an element of the subgroup $\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$ consisting of multiplication on the right by a unit quaternion, any two of these contact structures can be identified, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure identified in the first three cases.
Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.
This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.
For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.)
Let me address the first side question, i.e. Question: How many contact structures are there on odd-dimensional spheres?
An answer: Usually there are at least infinitely many inequivalent contact structures on $S^{2n+1}$.
1.The case of $S^{3}$:
The classification is complete only in the case of the three-sphere $S^3$ due to [Eliashberg][1]. There are essentially "countably many plus one" of them.
Here I talk about classification with respect to isotopies of contact structures: two contact structures are equivalent if there exists an isotopy of the underlying manifold sending one contact structure to the other. By Gray stability this is the same as existence of a homotopy of contact distributions from one to the other (the homotopy has to be through contact distibutions not just whatever distributions).
Now in dimension 3 there is a dichotomy -- either the contact manifold $(M, \xi)$ contains an overtwisted disc, in which case $\xi$ is called overtwisted, or not, in which case $\xi$ is called tight. By Eliashberg there exists up to isotopy a unique tight contact structure on $S^3$, the standard one. Moreover, every homotopy class of plane distributions on $S^3$ contains a unique overtwisted contact structure. There are $(\pi_{3}(S^2) \cong \mathbb{Z})$-many plane distributions on $S^3$ and so there are $\mathbb{Z}$-many inequivalent overtwised contact structures on $S^{3}$.
All in all, there $(\mathbb{Z}+1)$-many inequivalent contact structures on $S^3$.
2.Higher-dimensional spheres:
Only very recently a higher-dimensional analogue of an overtwisted disc has been discovered by [Borman-Eliasberg-Murphy][2] (this work has not yet been refereed, as far as I know). It states that every homotopy class of almost contact structures contains a unique (again up to isotopy) overtwisted contact struture.
An almost contact structure on a manifold is a co-dimension one distribution with a (linear) complex struture on it. In particular, the underlying distribution of a contact structure is an almost contact structure.
Homotopy classes of almost contact structures on $S^{2n+1}$ are classified by the homotopy group $\pi_{2n+1}(\mathrm{SO}(2n+2)/\mathrm{U}(n+1))$. In either the Ding-Geiges paper suggested by Fabrice Baudoin (in the first comment) or in the B-E-M paper [2] you can read that $$\pi_{2n+1}(\mathrm{SO}(2n+2)/\mathrm{U}(n+1)) = \begin{cases}\mathbb{Z}/n!\mathbb{Z}, & n=4k, \\ \mathbb{Z}, & n=4k+1, \\ \mathbb{Z}/\frac{n!}{2}\mathbb{Z}, & n=4k+2, \\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, & n=4k+3.\end{cases} $$
So there are at least that many inequivalent (overtwsited) contact structures on $S^{2n+1}$.
However, these are not necessarily all the contact structures. [Ustilovsky][3] constructed infinitely many inequivalent and non-overwisted contact structures on the spheres $S^{4m+1}$. So for example on $S^{5}$ there is a unique overtwisted contact structure but infinitely many non-overtwisted ones.
To conclude, you can watch Eliashberg's Banff talk on these recent discoveries (but it is not specifically about spheres).
[1]: Y.Eliasberg, Contact 3-manifolds twenty years since J.Martinet's work, link to the paper on Numdam.
[2]: M.S.Borman, Y.Eliashberg, E.Murphy, Existence and classification of overtwisted contact structures in all dimensions, arXiv.
[3]: I.Ustilovsky, Infinitely Many Contact Structures on $S^{4m+1}$, Internat. Math. Res. Notices 1999, no. 14, 781–791.
To answer your second side question, constructing a sphere as a homogeneous space is the same as saying that some Lie group acts transitively on it. The list of such is known as the Berger classification, and it coincides with the list of possible holonomy groups of Riemannian structures plus two extra cases. See http://en.wikipedia.org/wiki/Holonomy#The_Berger_classification
The list is quite short, so you can probably go through all cases and look for contact structures.