Question about Charge and Gauge Transformation

You are right of course! Physical observables must be gauge invariant. But this does not mean that they must be neutral. They could be charged under the global symmetry and be neutral under the local gauge symmetry.

In particular, a local gauge symmetry is generated by a function $\alpha(x)$ where $\alpha(x) \to 0$ as $|x| \to \infty$. A global symmetry of course has $\alpha(x) = $ constant which does not satisfy the above property. One way to have a charged gauge invariant operator is to connect it to a Wilson line that joins the operator to a point at infinity.

To add a bit more detail, a Wilson line $W_{{\cal P},q}(x_1,x_2)$ is a line operator (defined along a path ${\cal P}$) that under a gauge symmetry transforms as (assuming abelian symmetry for simplicity) $$ W_{{\cal P},q}(x_1,x_2) \to e^{- i q \alpha(x_1) } W_{{\cal P},q}(x_1,x_2) e^{ i q \alpha(x_2) } . $$ A charged local operator transforms under gauge symmetry as $$ {\cal O}(x) \to e^{ - i q \alpha(x) } {\cal O}(x) . $$ where $q$ is the charge of the state. We now construct the operator $$ {\tilde {\cal O}}(x) = W_{{\cal P},q}(\infty,x){\cal O}(x) $$ This transforms as $$ {\tilde {\cal O}}(x) \to e^{ - i q \alpha(\infty)} {\tilde {\cal O}}(x) . $$ Then, ${\tilde {\cal O}}(x)$ is invariant under local gauge transformations but not invariant under global symmetry transformations.