A parallel vector field implies a reduction of holonomy. Any form of $G_2$ does not preserve any nonzero vector when acting in its nontrivial 7-dimensional representation. So the holonomy must be the subgroup of $G_2$ preserving a nonzero vector, i.e. $SU(3)$. The reduction of holonomy group will, by deRham's splitting theorem, give the manifold a product structure, at least locally, so a local product of the line and a Calabi-Yau. The parallel vector field is dual to a parallel closed 1-form. If the manifold is compact, that implies that the first Betti number is nonzero. Moreover, applying the Cheeger-Gromoll splitting theorem, after taking a finite covering of the compact manifold, it splits into a product of a torus and a manifold of holonomy a subgroup of $G_2$. Strictly speaking, if you follow the definition in Wikipedia page (as in the question above), these products are actually $G_2$ manifolds, so the statement in the question, that there are no $G_2$ manifolds with parallel vector fields, is not correct. The correct statement is that the holonomy of a compact $G_2$ manifold is a proper subgroup of $G_2$ (up to conjugacy in $SO(7)$) if and only if the manifold has a finite Riemannian covering by a product of a torus and a compact manifold of dimension less than $7$ with holonomy a subgroup of $SU(3)$ (up to conjugacy in $G_2$).

For more information on the Cheeger--Gromoll splitting theorem, you might look at

- the expository paper: Cheeger,
*Structure theory and convergence in Riemannian geometry*, **Milan J. Math.** 78 (2010), no. 1, 221–264
- or Eschenburg and Heintze,
*An elementary proof of the Cheeger Gromoll splitting theorem*, **Annals of Global Analysis and Geometry**, June 1984, Volume 2, Issue 2, pp 141–151.
- or Jeff Cheeger and Detlef Gromoll,
*The splitting theorem for manifolds of nonnegative Ricci curvature*, **J. Differential Geom.**, Volume 6, Number 1 (1971), 119-128.