Well, let's consider what such a property would entail. $R$ is said to be antisymmetric iff $$\forall a,b\in A(a\:R\:b\:\wedge\:b\:R\:a\:\Rightarrow\: a=b).$$ Observe that if $p,q$ are statements, then $p\Rightarrow q$ is logically equivalent to $q\vee\neg p$, and its negation would be $p\wedge\neg q$.
In this case, by "opposite", it seems you're intending the opposed universal statement, rather than simply the negation. That is, we're only negating the latter part, e.g: $\forall p(q)$ becomes $\forall p(\neg q)$. Thus, $R$ is the "opposite" of antisymmetric iff $$\forall a,b\in A(a\:R\:b\:\wedge\:b\:R\:a\:\wedge\:a\neq b).$$ Consider that statement, though. The only set $A$ on which it could hold would be $A=\emptyset$. (Why?) Thus, only the empty relation is the "opposite" of antisymmetric. But the empty relation is also antisymmetric, so this isn't really a useful or particularly relevant property.
As to the latter part of your question, an equivalence relation can be antisymmetric. Consider the identity relation $R:=\bigl\{\langle a,b\rangle\in A\times A:a=b
\bigr\}$, for example. In fact, this is the only antisymmetric equivalence relation on $A$.
Proof: Let $S$ be an antisymmetric equivalence relation on $A$. It suffices to show that for all $a,b\in A,$ we have $a\:S\:b$ if and only if $a\:R\:b$. By reflexivity, we have that $a\:R\:b$ implies $a\:S\:b$. On the other hand, if $a\:S\:b$, then $b\:S\:a$ by symmetry, and then by antisymmetry we have $a\:R\:b$.