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. Indeed, fix a real number $a$, as an isotopy from the identity to $f^a$ you take the path $\varphi_{t}=f^{ta}$ with $t$ between $0$ and $1$. The vector field you call $\dot \varphi_{t}$ is just $aX$. The form $\iota_{aX}\omega$ is invariant by $\varphi_{t}$ so the integral you wrote to define the flux of $f^a$ is just the integral from $0$ to $1$ of the constant $(n-1)$-form $\iota_{aX}\omega$.
So the flux of $f^a$ is $a$ times the class of the form $\iota_{X}\omega$.
If the constant is nonzero, then for small $a$ 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 can take M to be any closed oriented surface of positive genus. If you pick any closed $1$-form $\alpha$, there always exists a volume preserving vector field $X_{\alpha}$ such that $\iota_{X_{\alpha}}\omega = \alpha$. So the cohomology class of $\iota_{X}\omega$ can be any class in $H^1$.
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).