Let $(M,J,\omega, \Omega)$ be a calabiyau manifold (not necessary compact). Does it follow that the holonomy group of $M$ is $SU_{n}$, where $n$ is the complex dimension of $M$ ?

Well, it depends on what you call a CalabiYau manifold (there are several possible terminologies indeed). First of all, a compact Kähler manifold with trivial canonical class does not necessarily have holonomy group $SU_n$ (with respect to some Ricciflat metric). More precisely, the holonomy group of some compact Kähler manifold $(X,\omega)$ is included in $SU_n$ iff there exists a nonzero parallel holomorphic $n$form. As a consequence the restricted holonomy group $H_0$ is included in $SU_n$ iff $(X,\omega)$ is Ricciflat. Now the holonomy groups of a Ricci flat compact Kähler manifold can be smaller than $SU_n$: think about any torus (the holonomy is trivial) or any holomorphic symplectic variety (the holonomy is $SP(n/2)$. However, there is a result, which was maybe what you had in mind: Theorem. Let $(X,\omega)$ be a compact Kähler manifold of dimension $n\geq 3$ with holonomy group $SU_n$. Then $X$ is projective and $H^0(X, \Omega_X^p)=0$ for every $0 < p < n$ and $\chi(\mathcal O_X)=1+(1)^n$. A manifold with such properties is sometimes called CalabiYau, indeed. For a reference, see Beauville's article "Variétés Kähleriennes à première classe de Chern nulle". As for the noncompact case, I don't know if such a results holds. 

