José is correct, with the caveat that Gunnar mentioned - you need simple-connectedness to know that reduced holonomy = holonomy. Below I expand a bit more on the details. [Thanks to Tim Perutz for catching errors in the initial version of this answer.]
Notice that the OP did not ask for $\Omega$ to be parallel or even closed. The following is true: If $(M, J, g, \omega)$ is Ricci-flat Kaehler, then the image of the first Chern class $c_1 (M)$in $H^2 (M, \mathbb R)$ vanishes, so that if $\pi_1(M) = 0$, then $H^2(M, \mathbb Z)$ has no torsion, and thus the canonical bundle $\Lambda^{n, 0} (M)$ is topologically trivial. So there exists a nowhere vanishing smooth $(n,0)$-form $\Omega$ that trivializes the canonical bundle. By consideration of type, $\Omega \wedge \overline \Omega$ is a nonvanishing $(n,n)$-form, so by rescaling $\Omega$ by a nowhere vanishing complex valued function, one gets for "free" the identity that
$$ \frac{\omega^n}{n!} = (-1)^{\frac{n(n-1)}{2}} \Omega \wedge \overline \Omega.$$
Since $\Omega$ is type $(n,0)$ and the complex structure is integrable, then $\Omega$ will be holomorphic (and thus the canonical bundle is holomorphically trivial) if and only if it is closed. Since $M$ is Ricci-flat, the Bochner theorem tells you that an $(n,0)$ form is closed if and only if it is parallel, which would give you holonomy contained in $SU(n)$.
Compactness is needed to go the other way: Yau's theorem says that if $M$ is compact Kaehler and $c_1 (M) = 0$, then there exists a unique Ricci flat Kaehler metric in each Kaehler class. There are noncompact examples where uniqueness fails. I don't know as much as I should about the literature on existence in the noncompact case, but the papers of Tian-Yau should have the answer.
A good elementary reference is Chapter 6 of Compact Manifolds with Special Holonomy by Dominic Joyce.
