If M is simply connected and closed (and oriented, but this is the case if it has a volume form), then $H^{1}(M)$ is zero and hence by Poincaré duality $H^{n-1}(M)$ is also zero. So the flux homomorphism is trivial and this answers your question.
Otherwise I think that the answer to your question is no. If you have a volume preserving vector field X on your manifold, with flow $f^t$ then the flux of $f^t$ is equal to a constant time t.
If the constant is nonzero, then for small t you get volume preserving diffeomorphisms arbitrarily close to the identity (in the $C^{\infty}$ topology) and with nonzero flux.
And you can find examples of pairs M, X like this. For instance you an take M to be any closed oriented surface of positive genus.
The condition to have zero flux is necessary to be able to write a diffeomorphism as a composition of diffeomorphisms compactly supported in open sets diffeomorphic to balls. (and isotopic to the identity inside these balls)
There is a recent book by Bounemoura (in french), about the same topics as Banyaga's book, but focusing mainly on surfaces, and with connections to more recent developments. You might find it interesting (it is called Simplicité des groupes de transformations de surfaces. [The simplicity of surface transformation groups] Ensaios Matemáticos [Mathematical Surveys], 14).