For finite sets the answer is no if $|B|>|A|\geq 2$.  Let $T_n$ be the full transformation monoid on $n$-letters (so the monoid of all maps on $n$-letters).  Let $C_n$ be the two-sided ideal of constant maps.  It is the minimal ideal of $T_n$.  Any non-injective homomorphism from $T_n$ must identify all the constant maps.  This is because $T_n$ acts faithfully on the left of the constant maps (assuming $T_n$ acts on the left of $1,\ldots, n$) (it looks just like the action of $T_n$ on $1,\ldots n$).  So if you identify no constant maps, then you must be injective.  If you identify some constant maps you will get a system of imprimitivity for the symmetric group $S_n$ acting on the left of $C_n$ and this action is primitive, being two-transitive.  So you must identify all the constant maps.  Thus every proper quotient of $T_n$ has a zero element.  Since $T_m$ only has a zero element when $m=1$ there is no surjective homomorphism $T_n$ to $T_m$ with $1<m<n$.

This argument shows that the unique minmal congruence on $T_n$ is to identify the constant maps.

**Added.**
I think the same argument should also handle the infinite case.  This is a little outside my comfort zone but I think for any set $A$ with at least two elements the symmetric group is two-transitive on $A$ and hence acts primitively and so any congruence on the full transformation monoid on $A$ will have to either be injective on constant maps or collapse them to a point and in the latter case, the image is not a full transformation monoid on more than one element.  In the former case, it must be injective on the full transformation monoid.  Since the set of constant maps is algebraically determined as the unique minimal two-sided ideal the cardinality of the set $A$ is determined.