If our Lagrangian is invariant under a local symmetry, then, by simply restricting our local symmetry to the case in which the transformation is constant over space-time, we obtain a global symmetry, and hence a corresponding Noether charge.
Because, however, this Noether charge didn't come from just any old symmetry, but in fact, a local symmetry, we might be able to say something special about it. In particular, I believe that the Noether charge should vanish, but I don't know why. If this is indeed the case, how do we prove it?
(Note that I don't want to make the assumption that local=gauge (i.e. non-physical).)
Best Answer
Assume that the Lagrangian density
$$\tag{1} {\cal L} ~=~ {\cal L}(\phi(x), \partial \phi(x), x) $$
does not depend on higher-order derivatives $\partial^2\phi$, $\partial^3\phi$, $\partial^4\phi$, etc. Let
$$\tag{2} \pi^{\mu}_{\alpha} ~:=~ \frac{\partial {\cal L}}{ \partial (\partial_{\mu}\phi^{\alpha})} $$
denote the de Donder momenta, and let
$$\tag{3} E_{\alpha}~:=~ \frac{\partial {\cal L}}{ \partial \phi^{\alpha}} - d_{\mu} \pi^{\mu}_{\alpha} $$
denote the Euler-Lagrange equations. Let us for simplicity assume that the infinitesimal local quasi-symmetry$^1$ transformation
$$\tag{4} \delta_{\varepsilon} \phi^{\alpha}~=~ Y^{\alpha}(\varepsilon) ~=~Y^{\alpha}\varepsilon + Y^{\alpha,\mu} d_{\mu}\varepsilon $$
is vertical$^2$ and that it does not depend on higher-order derivatives of the infinitesimal $x$-dependent parameter $\varepsilon$. [It is implicitly understood that the structure coefficients $Y^{\alpha}$ and $Y^{\alpha\mu}$ are independent of the parameter $\varepsilon$. If the theory has more that one symmetry parameter $\varepsilon^a$, $a=1, \ldots m$, we are just investigating one local symmetry (and its conservations law) at the time.] The bare Noether current $j^{\mu}(\varepsilon)$ is the momenta times the symmetry generators
$$\tag{5} j^{\mu}\varepsilon + j^{\mu,\nu}d_{\nu}\varepsilon ~=~j^{\mu}(\varepsilon) ~:=~ \pi^{\mu}_{\alpha}Y^{\alpha}(\varepsilon) ,$$
$$\tag{6} j^{\mu}~:=~ \pi^{\mu}_{\alpha}Y^{\alpha}, \qquad j^{\mu,\nu}~:=~ \pi^{\mu}_{\alpha}Y^{\alpha,\nu}. $$
(Again, it is implicitly understood that the structure coefficients $j^{\mu}$ and $j^{\mu\nu}$ are independent of the parameter $\varepsilon$, and so forth.) That the infinitesimal transformation (4) is a local quasi-symmetry$^1$ implies that variation of the Lagrangian density ${\cal L}$ wrt. (4) is a total space-time divergence
$$ d_{\mu} f^{\mu}(\varepsilon) ~=~ \delta_{\varepsilon} {\cal L} ~\stackrel{\begin{matrix}\text{chain}\\ \text{rule}\end{matrix}}{=}~ \frac{\partial {\cal L}}{ \partial \phi^{\alpha}} Y^{\alpha}(\varepsilon) + \pi^{\mu}_{\alpha}d_{\mu}Y^{\alpha}(\varepsilon) $$ $$\tag{7} ~\stackrel{\begin{matrix}\text{Leibniz'}\\ \text{rule}\end{matrix}}{=}~ E_{\alpha}Y^{\alpha}(\varepsilon) + d_{\mu} j^{\mu}(\varepsilon). $$
Here$^3$
$$ \tag{8} f^{\mu}(\varepsilon) ~=~ f^{\mu}\varepsilon + f^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} f^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon $$
are some functions with
$$\tag{9}f^{\mu,\nu\lambda}~=~f^{\mu,\lambda\nu}. $$
The full $\varepsilon$-dependent Noether current $J^{\mu}(\varepsilon)$ is defined as$^3$
$$\tag{10} J^{\mu}\varepsilon + J^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} J^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon ~=~J^{\mu}(\varepsilon) ~:=~ j^{\mu}(\varepsilon) - f^{\mu}(\varepsilon), $$
where
$$\tag{11}J^{\mu,\nu\lambda}~=~J^{\mu,\lambda\nu}. $$
Eqs. (7) and (10) imply the $\varepsilon$-dependent off-shell Noether identity
$$ \tag{12} d_{\mu} J^{\mu}(\varepsilon) ~=~ -E_{\alpha}Y^{\alpha}(\varepsilon) . $$
The $\varepsilon$-dependent off-shell Noether identity (12) is the key identity. Decomposing it in its $\varepsilon$-independent components leads to the following set (13)-(16) of identities,
$$ \tag{13} d_{\mu}J^{\mu} ~=~-E_{\alpha} Y^{\alpha} , $$
$$ \tag{14} J^{\mu} + d_{\nu} J^{\nu,\mu}~=~-E_{\alpha} Y^{\alpha,\mu} ,$$
$$ \tag{15} J^{\nu,\lambda}+J^{\lambda,\nu}+d_{\mu}J^{\mu,\nu\lambda} ~=~0 , $$
$$ \tag{16} \sum_{{\rm cycl}.~\mu,\nu,\lambda}J^{\mu,\nu\lambda} ~=~0, $$
in accordance with Noether's second theorem. Eq. (13) is just the usual off-shell Noether identity, which can be derived from the global symmetry alone via Noether's first theorem (where $\varepsilon$ is $x$-independent). As is well-known, the eq. (13) implies an on-shell conservation law
$$ \tag{17} d_{\mu}J^{\mu}~\approx~ 0, $$
or more explicitly written as
$$ \tag{18} \frac{d Q}{dt}~\approx~ 0,\qquad Q~:=~\int_{V} \! d^3V ~J^0. $$
(Here the $\approx$ sign denotes equality modulo Euler-Lagrange equations $E_{\alpha}\approx 0$. We have assume that the currents $J^i$, $i\in\{1,2,3\}$, vanish at the boundary $\partial V$.)
The remaining eqs. (14)-(16) may be repackaged as follows. Define the second Noether current ${\cal J}^{\mu}(\varepsilon)$ as$^4$
$$ \tag{19} {\cal J}^{\mu}\varepsilon + {\cal J}^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} {\cal J}^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon ~=~ {\cal J}^{\mu}(\varepsilon)~:= ~ J^{\mu}(\varepsilon)+ E_{\alpha} Y^{\alpha,\mu}\varepsilon. $$
It satisfies an $\varepsilon$-dependent off-shell conservation law
$$ d_{\mu} {\cal J}^{\mu}(\varepsilon) ~\stackrel{(12)+(19)}{=}~ -E_{\alpha}Y^{\alpha}(\varepsilon)+d_{\mu}(E_{\alpha} Y^{\alpha,\mu}\varepsilon)$$ $$ \tag{20}~\stackrel{(13)+(14)}{=}~ - \varepsilon d_{\mu}d_{\nu} J^{\nu,\mu}~\stackrel{(15)}{=}~\frac{\varepsilon}{2}d_{\mu}d_{\nu}d_{\lambda} J^{\lambda,\mu\nu}~\stackrel{(16)}{=}~0 . $$
One may introduce a so-called superpotential ${\cal K}^{\mu\nu}(\varepsilon)$ as$^3$
$$ {\cal K}^{\mu\nu}\varepsilon+{\cal K}^{\mu\nu,\lambda}d_{\lambda}\varepsilon~=~{\cal K}^{\mu\nu}(\varepsilon)~=~-{\cal K}^{\nu\mu}(\varepsilon) $$ $$~:=~ \left(\frac{1}{2} J^{\mu,\nu}-\frac{1}{6}d_{\lambda}J^{\mu,\nu\lambda}\right)\varepsilon+ \frac{1}{3} J^{\mu,\nu\lambda}d_{\lambda}\varepsilon-(\mu\leftrightarrow \nu)$$ $$ \tag{21}~\stackrel{(14)+(16)}{=}~ \left( J^{\mu,\nu}+\frac{1}{3}d_{\lambda}(J^{\lambda,\mu\nu}-J^{\mu,\nu\lambda})\right)\varepsilon+ \frac{1}{3}\left( J^{\mu,\nu\lambda}-J^{\nu,\mu\lambda}\right)d_{\lambda}\varepsilon$$
A straightforward calculation
$$ d_{\nu}{\cal K}^{\mu\nu}(\varepsilon) ~\stackrel{(15)+(21)}{=}~J^{\mu,\nu}d_{\nu}\varepsilon -\varepsilon d_{\nu}\left(J^{\nu,\mu}+d_{\lambda}J^{\lambda,\mu\nu}\right)$$ $$ \tag{22}+\frac{\varepsilon}{3}d_{\nu}d_{\lambda}\left(J^{\lambda,\mu\nu}-J^{\mu,\nu\lambda}\right) +\frac{1}{3}\left( J^{\mu,\nu\lambda}-J^{\nu,\mu\lambda}\right)d_{\nu}d_{\lambda}\varepsilon ~\stackrel{(14)+(16)+(19)}{=}~{\cal J}^{\mu}(\varepsilon)$$
shows that ${\cal K}^{\mu\nu}(\varepsilon)$ is the superpotential for the second Noether current ${\cal J}^{\mu}(\varepsilon)$. The existence of the superpotential ${\cal K}^{\mu\nu}(\varepsilon)=-{\cal K}^{\nu\mu}(\varepsilon)$ makes the off-shell conservation law (20) manifest
$$ \tag{23}d_{\mu}{\cal J}^{\mu}(\varepsilon)~\stackrel{(22)}{=}~d_{\mu}d_{\nu}{\cal K}^{\mu\nu}(\varepsilon)~=~0. $$
Moreover, as a consequence the superpotential (22), the corresponding second Noether charge ${\cal Q}(\varepsilon)$ vanishes off-shell
$$ \tag{24}{\cal Q}(\varepsilon)~:=~\int_{V} \! d^3V ~{\cal J}^0(\varepsilon) ~=~\int_{V} \! d^3V ~d_i{\cal K}^{0i}(\varepsilon) ~=~\int_{\partial V} \! d^2\!A_i ~{\cal K}^{0i}(\varepsilon)~=~0, $$
if we assume that the currents ${\cal J}^{\mu}(\varepsilon)$, $\mu\in\{0,1,2,3\}$, vanish at the boundary $\partial V$.
We conclude that the remaining eqs. (14)-(16) are trivially satisfied, and that the local quasi-symmetry doesn't imply additional non-trivial conservation laws besides the ones (13,17,18) already derived from the corresponding global quasi-symmetry. Note in particular, that the local quasi-symmetry does not force the conserved charge (18) to vanish.
This is e.g. the situation for gauge symmetry in electrodynamics, where the off-shell conservation law (20) of the second Noether current ${\cal J}^{\mu}=- d_{\nu}F^{\nu\mu}$ is a triviality, cf. also this and this Phys.SE posts. Electric charge conservation follows from global gauge symmetry alone, cf. this Phys.SE post. Note in particular, that there could be a nonzero surplus of total electric charge (18).
--
$^1$ An off-shell transformation is a quasi-symmetry if the Lagrangian density ${\cal L}$ is preserved $\delta_{\varepsilon} {\cal L}= d_{\mu} f^{\mu}(\varepsilon)$ modulo a total space-time divergence, cf. this Phys.SE answer. If the total space-time divergence $d_{\mu} f^{\mu}(\varepsilon)$ is zero, we speak of a symmetry.
$^2$ Here we restrict for simplicity to only vertical transformations $\delta_{\varepsilon} \phi^{\alpha}$, i.e., any horizontal transformation $\delta_{\varepsilon} x^{\mu}=0$ are assumed to vanish.
$^3$ For field theory in more than one space-time dimensions $d>1$, the higher structure functions $f^{\mu,\nu\lambda}=-J^{\mu,\nu\lambda}$ may be non-zero. However, they vanish in one space-time dimension $d=1$, i.e. in point mechanics. If they vanish, then the superpotential (21) simplifies to ${\cal K}^{\mu\nu}(\varepsilon)=J^{\mu,\nu}\varepsilon$.
$^4$ The second Noether current is defined e.g. in M. Blagojevic and M. Vasilic, Class. Quant. Grav. 22 (2005) 3891, arXiv:hep-th/0410111, subsection IV.A and references therein. See also Philip Gibbs' answer for the case where the quasi-symmetry is a symmetry.