I believe the answer is that the class of an element of $g \in A_n$ becomes a single class when lifted to the 2-fold central covering group of $A_n$ if and only if $g$ has even order and $g$ has (at least) two cycles of the same length (including cycles of length 1).
The class of an element of $g \in S_n$ lifts to a single class in either of the two covering groups of $S_n$ if and only if either the condition for $g \in A_n$ holds, or $g$ has an even number of cycles of even length.
Let $\hat{S}$ and $\hat{A}$ be covering groups of $S_n$ and $A_n$. Let $z$ be the central element of order 2 in $\hat{S}$. To avoid too many hats, I will denote the inverse image in $\hat{S}$ of $g \in S_n$ also by $g$.
I believe that to study these covering groups, all you really need to know is that if $g$ and $h$ are disjoint transpositions in $S_n$, such as $(1,2)$ and $(3,4)$, then $[g,h] = z$ in $\hat{S}$. Also, for two commuting pairs of transpositions in $A_4$, like $g=(1,2)(3,4)$ and $h = (1,3)(2,4)$, we have $[g,h]=z$ in $\hat{A}$.
Clearly, if $g$ has odd order $k$ in $A_n$ or $S_n$, then its class lifts to two classes in $\hat{A}$ or $\hat{S}$, one of order $k$ and one of order $2k$, so we only need consider elements $g$ of even order.
If $g$ has two cycles of equal even length, such as $(1,2,3,4)(5,6,7,8)$ (+ other cycles) then we can take $h = (1,5)(2,6)(3,7)(4,8)$ and get $[g,h] = z$ in $\hat{A}$, so the class of $g$ lifts to a single class in $\hat{A}$. I had some difficulty proving that, but if it was false, then you could show that the centralizer of $h$ in $\hat{A}$ mapped on to the full centralizer of $h$ in $A_n$, which is not true, because the image of the centralizer does not contain $(1,2)(5,6)$.
On the other hand, if $g$ has even order and has two cycles of the same odd length, such as $g = (1,2)(3,4,5,6)(7,8,9)(10,11,12)$ with $n=12$, then we can take $h = (1,2)(7,10)(8,11)(9,12)$ and get $[g,h] = z$ in $\hat{A}$.
But if all cycles of $g$ have different lengths, then elements $h$ of the centralizer of $g$ in $A_n$ are made up of powers of those cycles, and since $h$ must have an even number of cycles of even length, we always get $[g,h]=1$ in $\hat{A}$.
However, in $S_n$, if $g$ has an even number of cycles of even length, like $g=(1,2)(3,4,5,6)$ with $n=6$, then we can take $h=(1,2)$ and get $[g,h]=z$ in $\hat{S}$.
Let me know if I have got something wrong, or if you would like more detail!.
Best Answer
What you need is an example of a subgroup $G\subset A_n$ such that the normalizer of $G$ in $S_n$ is contained in $A_n$. If every automorphism of $G$ is inner, then it will be enough if the centralizer of $G$ in $S_n$ is contained in $A_n$. How about $n=8$ and $G$ the diagonal copy of $S_4$ in $S_4\times S_4\subset S_8$, if you know what I mean?
EDIT: If you replace this $G$ by its normal subgroup of order $4$, you get another example. This one is minimal with regard to the order of $G$ (though not with regard to $n$, as Derek's Holt's examples show).