TheI'm assuming that "minimal generating set" means "a generating set that does not properly contain a generating set" (that is, every proper subset does not generate). In that reading, the statement is false in both finite and infinite ranks.
For a counterexample in finite rank, let $G$ be infinite cyclic generated by $x$ (rank $1$), and $H$ be cyclic of order $2$ generated by $y$ (also rank $1$). Let $f\colon G\to H$ be the map sending $x$ to $y$. The minimal generating set $\{x^2,x^3\}$ (which generates $G$, but no proper subset does) is mapped to $\{e,y\}$, which is not minimal. Or you could take $H$ to be cyclic of order $4$, if you want to avoid the trivial element, as then you get $y^2$ and $y^3$; which is not minimal.
For infinite rank, take $G=H$ be a direct sum of countably infinitely many copies of the infinite cyclic group; take $\{x_i\}_{i=1}^{\infty}$ as a basis for $G$, and $\{y_j\}_{j=1}^{\infty}$ as a basis for $H$. Take $f\colon G\to H$ be the map that sends $x_{2i-1}$ to $y_i$ and $x_{2i}$ to $y_i^{-1}$.
If by "minimal generating set" you mean "generating set of minimal cardinality", then given that the rank of a group is the least cardinality of a generating set, then the answer is positive: under a surjective homomorphism, a generating set is mapped to a generating set. So if $X$ is a generating set for $G$ with $\mathrm{rank}(G)$ elements, then $f(X)$ is a generating set for $H$ with at most $\mathrm{rank}(G)=\mathrm{rank}(H)$ elements, and since any generating set has at least $\mathrm{rank}(H)$ elements, it follows that $f(X)$ has exactly $\mathrm{rank}(H)$ elements (in cardinality) and so is a generating set of minimal cardinality.