My formulation would be:
There are no functors from the category $\mathcal{V}$ of real vector spaces( where morphisms are linear maps) into the category $\mathcal{V}_b$ of real vector spaces with a "distinguished" base satisfying $$F(V)=(V,B)$$ for all objects $ V \in \mathcal{V}.$ In this statement the objects of $\mathcal{V}_b$ are ordered pairs $(V,B)$ with $V$ a vector space and $B$ a base in $V.$ The morphisms between $(V,B)$ and $(W,B')$ are linear maps $f:V\to W$ such that $f(B) \subset B'.$
Your formulation is too strict, as in the positive case when we take the full subcategory of $K^n$ (where $K$ is the base field), there should be such a functor, but linear transformations don't preserve the standard basis in general.
Actually, using axiom of choice, we can equip each (finite dimensional) vector space $V$ simultaneously by a fixed basis $B$ and thus obtain a functor $V\mapsto (V,B)$ (which doesn't necessarily satisfy your criteria) and that can be further enhanced to an equivalence to the above subcategory, or to the category whose arrows are matrices (with entries from $K$).