These two statements are not contradicting. Charge is conserved regardless of the symmetry being broken or being not broken.
The usual wisdom goes like this: assume you have a conserved current $j^\mu(x)$ with
$$ \partial_\mu j^\mu = 0 \ . $$
Now pick a constant-time ball $\Sigma$ and define
$$ Q = \int_\Sigma j^0(x) d^3 x \ . $$
This exists if the charge density decays fast enough. This is a conserved charge since:
$$ \frac{d Q}{d t} = \int_\Sigma \partial_0 j^0(x) d^3 x = - \int_\Sigma \nabla \cdot j(x) d^3 x = - \int_{\partial \Sigma} j(x) \cdot d \sigma $$
Where in the last equality we used that the currents decay at infinity.
Now this is classical reasoning. On the quantum level, the current becomes an operator, and all the statements about the fall-off of functions at infinity are statements about the states we wish to apply these operators on. In particular, it suffices to look at the vacuum, since we usually only consider states which differ from the vacuum in a finite region, so once the vacuum is fine, all excited states will be fine as well.
But in the case of spontaneous symmetry breaking, it is just not true that
$$ j^0(x)|0\rangle $$
decays at infinity, hence the integral defining the charge does not exist. Note however that while $Q$ doesn't make sense in itself, one may still give meaning to commutators of $Q$ with field operators $\phi(x)$:
$$ [ Q, \phi(x) ] $$
And in particular, these still implement the symmetry.
