Is there any necessary and sufficient condition for existence of $SU(3)$structure on 6manifolds $M$?

Yes, it is wellknown that a $6$manifold has an $\mathrm{SU}(3)$structure if and only if it is orientable and spinnable (i.e., it has a spin structure). The necessity of these two conditions is clear, since $\mathrm{SU}(3)$ is both connected and simplyconnected. For the sufficiency, first note that if $M^6$ is orientable and spinnnable (i.e., the first two StiefelWhitney classes of its tangent bundle vanish), then, fixing a Riemannian metric and orientation on $M$, one can choose a spin structure and construct the corresponding spinor bundle $\mathbb{S}\to M$. Since $\mathrm{Spin}(6)\simeq\mathrm{SU}(4)$, this bundle is a complex $4$plane bundle, which is, of course, a real $8$plane bundle. Since $8>6$, there is a nonvanishing section of $\mathbb{S}$ over $M$, and this will reduce the structure group of this bundle from $\mathrm{SU}(4)$ to $\mathrm{SU}(3)$. This will, correspondingly, reduce the structure group of the tangent bundle to $\mathrm{SU}(3)$ (since this subgroup does not contain $I_4\in\mathrm{SU}(4)$). Thus, $M$ carries an $\mathrm{SU}(3)$reduction of the structure group of the tangent bundle. NB: This is a classical argument, but I don't know who first wrote it down. Maybe, Alfred Gray did it first, but I am not sure, and I will not have any access to reference books for the next 12 days while I am traveling, so I won't be able to check or give you further information. 

