My first idea was to argue that if there was a symplectic form $\omega$, then we would have $[\omega\wedge\omega]=[0]$ (in deRham cohomology), since $H_{dR}^2(\mathbb{R}^3\times\mathbb{S^3})=0$, which is absurd.
But I've noticed this is wrong, since the argument "$\omega\in\Omega^2(M)$ symplectic $\Rightarrow H_{dR}^2(M)\neq 0$" is valid only if $M$ is compact, which is not the case for $\mathbb{R}^3\times\mathbb{S}^3$.
Now I don't really know what to do.
Is there some generic strategy for this kind of problem?
I am shooting wildly here, but $S^3$ is parallelizable (it is a principal homogenous space for $\text{SU}(2)$), hence its tangent (and thus, cotangent) bundle is trivial and is diffeomorphic to $S^3\times\mathbb R^3$.
Cotangent bundles carry a natural symplectic form, hence $S^3\times\mathbb R^3$ admits symplectic forms.