If $r=2$ then $r'>0$. For an example where $r'=1$, take a curve such that the jacobian has a nontrivial endomorphism $f$ and such that the group of points in the jacobian is generated by $P,f(P)$ for some point $P$ Now find a prime $p$ splitting in $\mathbb{Q}(f)$ so that $f(P)=\alpha P$ for some $p$-adic number $\alpha$. Then $r'=1$.

Added later: In your toy example, the endomorphism ring contains the fifth roots of unity so it may fall in my example above for those primes that split in that ring. A reasonable conjecture would be that if the endomorphism ring of the jacobian is $\mathbb{Z}$, then $r' = \min \{ g,r \}$. This is likely to be very hard to prove, as it is an abelian variety analogue of Leopoldt's conjecture.

If $r=2$ then $r'>0$. For an example where $r'=1$, take a curve such that the jacobian has a nontrivial endomorphism $f$ and such that the group of points in the jacobian is generated by $P,f(P)$ for some point $P$ Now find a prime $p$ splitting in $\mathbb{Q}(f)$ so that $f(P)=\alpha P$ for some $p$-adic number $\alpha$. Then $r'=1$.

Added later: In your toy example, the endomorphism ring contains the fifth roots of unity so it may fall in my example above for those primes that split in that ring. A reasonable conjecture would be that if the endomorphism ring of the jacobian is $\mathbb{Z}$, then $r' = \min \{ g,r \}$. This is likely to be very hard to prove, as it is an abelian variety analogue of Leopoldt's conjecture.

Added much, much later: There are problems with my construction and what I say is probably not right. M.D. had a good reason to be skeptical in the comments. This was pointed out to me by Bjorn Poonen in a recent email exchange. His question is discussed further in a paper of M. Waldschmidt On the p-adic closure of a subgroup of rational points on an Abelian variety, Afr. Mat. 22 (2011), no. 1, 79–89.

If $r=2$ then $r'>0$. For an example where $r'=1$, take a curve such that the jacobian has a nontrivial endomorphism $f$ And such that the group of points in the jacobian is generated by $P,f(P)$ for some point $P$ Now find a prime $p$ splitting in $\mathbb{Q}(f)$ so that $f(P)=\alphaP$ for some $p$-adic number $\alpha$. Then $r'=1$.

If $r=2$ then $r'>0$. For an example where $r'=1$, take a curve such that the jacobian has a nontrivial endomorphism $f$ and such that the group of points in the jacobian is generated by $P,f(P)$ for some point $P$ Now find a prime $p$ splitting in $\mathbb{Q}(f)$ so that $f(P)=\alpha P$ for some $p$-adic number $\alpha$. Then $r'=1$.

Added later: In your toy example, the endomorphism ring contains the fifth roots of unity so it may fall in my example above for those primes that split in that ring. A reasonable conjecture would be that if the endomorphism ring of the jacobian is $\mathbb{Z}$, then $r' = \min \{ g,r \}$. This is likely to be very hard to prove, as it is an abelian variety analogue of Leopoldt's conjecture.