Let $\Gamma$ be a Fuchsian Group (that means a Discrete Subgroup of $RSL_2(\mathbb{R})$) acting on the Closed Unit Disc $\mathbb{D}$.
Let $\Lambda (\Gamma)$ be the Limit Set of $\Gamma$, i.e., it is the smallest closed subest of $\mathbb{D}$ that is invariant under the action of $\Gamma$.
Is there any theorem that states that if $\Gamma$ is Non-elementary (that means the limit set $\Lambda (\Gamma)$ contains at least three elements) then there is no invariant measure for the action of $\Gamma$ on its Limit Set $\Lambda (\Gamma)$.
First of all, your definition of the limit set $\Lambda(\Gamma)$ is not quite correct. For instance, you forgot to add nonempty. But also, your definition does not work for elementary subgroups. The standard definition is that the limit set is the accumulation set in $\partial {\mathbb D}$ of an orbit of $\Gamma$ in the interior of ${\mathbb D}$.
Now, to your question. The answer depends on your assumptions about a measure. A measure on a set $X$ consists of choice of a sigma-algebra ${\mathcal A}\subset 2^X$ and a function $\mu: {\mathcal A}\to [0,\infty)$ satisfying axioms of a measure.
For $X=\Lambda(\Gamma)$, let's take ${\mathcal A}\subset 2^X$ equal $\{\emptyset, X\}$ and $\mu(X)=1$, $\mu(\emptyset)=0$. This is clearly a measure and it is obviously $\Gamma$-invariant.
At the other extreme, suppose that $\mu$ is a Borel probability measure on $X=\Lambda(\Gamma)$. Then it is a pleasant exercise in using density in $X$ of fixed points of hyperbolic elements of $\Gamma$ to see that $\mu$ cannot be $\Gamma$-invariant (if $\Gamma$ is nonelementary, of course). If you want a hint for solving this exercise, use the following property:
If $\gamma\in \Gamma$ is hyperbolic with repulsive (resp. attractive) fixed points $\lambda_-, \lambda_+$ and $J$ is an open interval in $S^1$ containing $\lambda_-$, then $$ \bigcup_{n\ge 0} \gamma^n(J)= S^1 \setminus \{\lambda_+\}. $$