The answer is yes, and it's fairly elementary. By the usual 2-descent, the curve $C$ gives a class $c$ in $H^1(\mathbb{Q},E[2])$, where $E$ is the Jacobian you wrote down. As you vary $d$, the groups $H^1(\mathbb{Q},E_d[2])$ are canonically isomorphic, and $c$ is also the class of $C_d$.  To answer your question, you should condition on the possibility that $c$ "comes from" the 2-torsion subgroup of $E$ via $E(\mathbb{Q})/2E(\mathbb{Q}) \to H^1(\mathbb{Q},E[2])$. I don't know what the conditions on $a,b$ for this to happen are, but it won't matter.

If $C$ doesn't come from 2-torsion, and if $C_d$ has a rational point, then automatically $E_d$ has rank at least one (by descent). And as Alex B. says, it's easy to see that there are infinitely many squarefree $d$ for which this happens (his argument doesn't quite show this, but it's not hard to fiddle with congruence conditions to produce infinitely many squareclasses).   

If $C$ does come from 2-torsion, then $C_d$ has a rational point for all $d$, but there is no guarantee that the rank is at least 1. So your question reduces to asking about the ranks of the twists of $E$. But now just apply the argument from the previous paragraph to some other class in $H^1(\mathbb{Q},E[2])$ represented by a binary quartic form (there are infinitely many of these).