Is this distribution completely non integrable? We consider the usual Riemannian metric on $S^{n}$. Its corresponding LC connection gives us  a distribution on $TS^{n}$. Is this distribution completely nonintegrable?
In general, what type of curvature criterions can be used to prove that certain Levi Civita connection gives a completely non integrable distributation?
 A: Here is a more complete answer to your question, which describes exactly which vectors in $TM$ can be joined by $L$-curves.  (I changed $H$ to $L$ since I have to use $H$ for holonomy.)
Recall that, when $(M,g)$ is a Riemannian manifold (assumed connected and simply connected for simplicity), there is a unique horizontal plane field $L$ on $TM$ (i.e., $L$ is a complement to the kernel of $\pi':T(TM)\to TM$) with the property that a curve $\gamma:[a,b]\to TM$ is tangent to $L$ if and only if it is the $\nabla^g$-parallel translation of $\gamma(a)\in T_{\pi(\gamma(a))}M$ along $\pi\circ\gamma:[a,b]\to M$, where $\nabla^g$ is the Levi-Civita connection of $g$.
Fix a point $x\in M$ and let $H_x\subset\mathrm{SO}(T_xM,g_x)$ be the holonomy group of $\nabla^g$ at $x$.  Any $H_x$-invariant function $p_x:T_xM\to\mathbb{R}$ then extends uniquely to a function $p:TM\to\mathbb{R}$ that is constant along all $L$-curves.  For example, the function $g_x:T_xM\to\mathbb{R}$ extends uniquely to $g:TM\to\mathbb{R}$, and this function is constant along all $L$-curves, since $\nabla^g$-parallel translation preserves the $g$-norm of tangent vectors.
Conversely, it is not hard to show, using Berger's classification of Riemannian holonomy groups, that, if two vectors $v\in T_xM$ and $w\in T_yM$ have equal $p$-values for all such "$L$-constant" functions $p:TM\to\mathbb{R}$, then there is an $L$-curve that joins them.
To see why, remember that there is a decomposition $T_xM=V_1\oplus\cdots\oplus V_m$ into nontrivial subspaces $V_i$ for $1\le i\le m$ and a product decomposition
$$
H_x = H_1\times \cdots \times H_m
$$
where $H_i$ acts trivially on $V_j$ for $j\not=i$ and irreducibly on $V_i$.  Moreover, each $H_i$ is isomorphic to the holonomy group of an irreducible Riemannian manifold $(M_i,g_i)$ of dimension $d_i = \dim V_i\ge 1$, where the isomorphism is induced via some isometric isomorphism $T_{x_i}M_i\simeq V_i$.
Now, by Berger's classification, either $H_i$ acts transitively on the unit sphere in $V_i$ (in which case the ring of $H_i$-invariant polynomials is generated by $g_i$ and so has algebraic dimension $r_i=1$) or else $(M_i,g_i)$ is a Riemannian symmetric space of rank $r_i\ge1$.  In either case, there exists a ring $R_i$ of $H_i$-invariant polynomials on $V_i$ with algebraic dimension $r_i$ such that the polynomial functions in $R_i$ have the $H_i$-orbits as their simultaneous level sets.
Putting all of this together, we see that there is a finitely generated ring $R_x$ of $H_x$-invariant polynomials on $T_xM$ such that two vectors $v$ and $w$ in $T_xM$ can be joined by an $L$-curve if and only if they have the same $p_x$-value for all $p_x\in R_x$.  
In the specific case of $S^n$, it is known that, for any Riemannian metric $g$ on $S^n$, the holonomy group is $H_x = \mathrm{SO}(T_xM,g_x)$ for all $x\in S^n$.  Thus, when $n>1$, it follows that any two tangent vectors to $S^n$ can be joined by an $L$-curve if and only if they have the same $g$-length.
