Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered, oriented, closed 3-manifold $M$, with a fiber of genus $\ge 2$ and with pseudo-Anosov monodromy, and with 2nd homology of rank $\ge 2$, gives counterexamples.

The Thurston norm on $H_2(M;\mathbb{R})$ has a polyhedral unit ball, and there is a symmetric set of top dimensional faces, called "fibered faces", such that a homology class contained in the integer lattice $H_2(M;\mathbb{Z})$ of $H_2(M;\mathbb{R})$ is represented by a fiber of a fibration over $S^1$ if and only if that class is in the interior of the cone of a fibered face. Furthermore, inside the cone on a fibered face, the value of the norm is given by a rationally defined linear functional $x : H_2(M;\mathbb{R}) \to \mathbb{R}$ such that if $S$ is in the integer point in that cone then $x(S)=-\chi(S)$.

We are supposing that $M$ fibers and that $\text{rank}(M) \ge 2$, and so there exists a fibered face, with corresponding linear functional $x$. Consider nonempty integer level sets of the form $L_k = x^{-1}(k) \cap H_2(M;\mathbb{Z})$ ($k$ must be even for $L_k$ to be nonempty). As $k \to +\infty$, clearly the cardinality of $L_k$ gets larger and larger. Thus one obtains an arbitrarily large number of fibers all of the same Euler characteristic, but representing pairwise distinct homology classes.

But $M$ is a closed hyperbolic 3-manifold and hence its group of homeomorphisms modulo isotopy of $M$ is finite, say of cardinality $A$. So at most $A$ homology classes can represent conjugate mapping classes.