Does the inverse of a one-way function $f$ being reducible to a predicate $b$ imply that $b$ is a hard-core bit for $f$?
Before addressing your question, we should note that there are some subtleties when defining "inverting $f(x)$ reduces to finding $h(x)$". Suppose we aim towards something along the lines of a standard Turing reduction, then what does our oracle do? When feeding the oracle some string $y\in\{0,1\}^*$, where the only promise about $y$ is that it lies in the image of $f$, then what do you expect it to return? If $f$ is a permutation, then this is easy. However, when $f$ is not one to one, it isn't clear what to expect. One possibility is to require that the oracle returns $h(x)$ for some $x\in f^{-1}(y)$, though it might make the oracle pretty much useless (I'd guess that there exists a one way function $f$ and a hardcore predicate $h$ for $f$ such that any function in $2^{\{0,1\}^*}$ is an $h$-oracle for $f$ under this definition).

Back to your original question. Suppose $f$ is a one way permutation and define $g(x,i,j)=(f(x),i,j,x_i)$. Inverting $f$ reduces to (under the above definition) evaluating $h(x,i,j)=x_j$, since you can feed the oracle different values of $j$. However, $h$ is not hardcore for $g$, since for a uniformly chosen triplet $(x,i,j)$ where $x\in\{0,1\}^n$ and $i,j\in [n]$, with probability $\frac{1}{n}$ it holds that $i=j$ and you can recover $x_j$.