The answer is yes, provided ZF itself is consistent. 

The reason is that the existence of the hyperreals, in a context with the [transfer principle](http://en.wikipedia.org/wiki/Transfer_principle#Transfer_principle_for_the_hyperreals), implies that there is a nonprincipal ultrafilter on $\mathbb{N}$, because if $N$ is any nonstandard (infinite) natural number, then the set $U$ consisting of all $X\subset\mathbb{N}$ with $N\in X^*$ is a nonprincipal ultrafilter on $\mathbb{N}$:

 - If $X\in U$ and $X\subset Y$, then $N\in X^*\subset Y^*$, and so $Y\in U$. 
 - If $X,Y\in U$, then $n\in X^*=Y^*=(X\cap Y)^*$ and so $X\cap Y\in U$. 
 - If $X\subset\mathbb{N}$, then every number is in $X$ or in $\mathbb{N}-X$, and so either $N\in X^*$ or $N\in(\mathbb{N}-X)^*$ and thus $X\in U$ or $\mathbb{N}-X\in U$. 
 - For any particular standard natural number $n$, the set $X=\{m\in \mathbb{N}\mid n\leq m\}$ is in $U$, because $n^*\leq N$. 
So 

Basically, the ultrafilter concentrates on sets that express all and only the properties held by the nonstandard number $N$. (See also my answer to [A remark of Connes](http://mathoverflow.net/a/57108/1946), where I make a similar point.) 

Thus, in a model of ZF with no nonprincipal ultrafilter on $\mathbb{N}$, there is no structure of the hyperreals satisfying the transfer principle.