I think that the answer is yes and that we don't really need the full Whitney's trick for this since being homologous is a much coarser relation than being isotopic, which is what Whitney's trick gives. So, rather than using the full trick, one can use just a half of it.
Let $M$ be an oriented smooth manifold, and let $Z_1,Z_2$ be cycles represented by pseudo-manifolds (recall that a pseudo-manifold is a stratified space that has no codimension 1 strata; any homology class can be represented by a pseudo-manifold). Assume $\deg Z_2\geq 1$ (when one of the cycles is 0-dimensional the statement we're after is clearly true).
First, let's make $Z_2$ connected by joining the connected components with tubes. While doing this we may introduce new intersection points, but after a small isotopy these will be all transversal and their signs will add up to 0. Second, take two intersection points $P,Q$ with different signs and join them with a path $\gamma$ in $Z_2$.
Now let's modify $Z_1$ by taking out two small balls around $P$ and $Q$ in $Z_1$and inserting a thin tube around $\gamma$ instead. The result will be homologous to $Z_1$: the fact that $P$ and $Q$ have different signs ensures that the small balls around them and the tube together form the boundary of a tubular neighborhood of $\gamma$ in $Z_1$. [I wish I could draw a picture here but don't know how to do that.] Notice that when $Z_1$ is a loop around $(0,0)$ and $(1,0)$ and $Z_2$ is a loop around $(-1,0)$ and $(0,0)$, as in Simon Rose's example then this procedure cuts $Z_1$ into two loops, one around $(0,0)$ and the other around $(1,0)$.
In this way one can eliminate every pair of intersection points with opposite signs. Notice that we haven't done anything to $Z_2$ in the process, apart from making it connected.