I) Let there be given a local action functional

$$\tag{1} S[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}, $$

with the Lagrangian density

$$\tag{2} {\cal L}(\phi(x),\partial\phi(x),x). $$

[We leave it to the reader to extend to higher-derivative theories. See also e.g. Ref. 1.] 

II) We want to study an infinitesimal variation$^1$ 

$$\tag{3} \delta x^{\mu}~=~\epsilon X^{\mu} \qquad\text{and}\qquad
\delta\phi^{\alpha}~=~\epsilon Y^{\alpha}$$  

of spacetime coordinates $x^{\mu}$ and fields $\phi^{\alpha}$, with arbitrary $x$-dependent infinitesimal $\epsilon(x)$, and with some given fixed generating functions

$$\tag{4}  X^{\mu}(x)\qquad\text{and}\qquad Y^{\alpha}(\phi(x),\partial\phi(x),x).$$ 

Then the corresponding infinitesimal variation of the action $S$ takes the form$^2$

$$\tag{5} \delta S ~\sim~ \int_V \mathrm{d}^n x \left(\epsilon ~ k  +  j^{\mu} ~ d_{\mu} \epsilon \right) $$

for some structure functions 

$$\tag{6} k(\phi(x),\partial\phi(x),\partial^2\phi(x),x)$$ 

and 

$$\tag{7} j^\mu(\phi(x),\partial\phi(x),x).$$

[One may show that some terms in the $k$ structure function (6) are proportional to eoms, which are typically of second order, and therefore the $k$ structure function (6) may depend on second-order spacetime derivatives.]

III) Next we assume that the action $S$ has a quasisymmetry$^3$ for $x$-independent infinitesimal $\epsilon$. Then eq. (5) reduces to

$$\tag{8}  0~\sim~\epsilon\int_V \mathrm{d}^n x~ k. $$

IV) Now let us return to OP's question. Due to the fact that eq. (8) holds for all off-shell field configurations, we may show that eq. (8) is only possible if 

$$\tag{9} k ~=~ d_{\mu}k^{\mu} $$ 

is a total divergence. (Here the words _on-shell_ and _off-shell_ refer to whether the eoms are satisfied or not.) In more detail, there are two possibilities:

1.  If we know that eq. (8) holds for _every_ integration region $V$, we can deduce eq. (9) by localization. 

2. If we only know that eq. (8) holds for a _single fixed_ integration region $V$, then the reason for eq. (9) is that the Euler-Lagrange derivatives of the functional $K[\phi]:=\int_V \mathrm{d}^n x~ k$ must be identically zero. Therefore $k$ itself must be a total divergence, due to an algebraic Poincare lemma of the so-called bi-variational complex, see e.g. Ref. 2. [Note that there could in principle be topological obstructions in field configuration space which ruin this proof of eq. (9).] See also [this](https://physics.stackexchange.com/a/22282/2451) related Phys.SE answer by me. 

V) One may show that the $j^\mu$ structure functions (7) are precisely the bare Noether currents. Next define the full Noether currents 

$$\tag{10} J^{\mu}~:=~j^{\mu}-k^{\mu}.$$ 

On-shell, after an integration by part, eq. (5) becomes 

$$\tag{11}  0~\approx~ \delta S 
~\stackrel{(5)+(9)+(10)}{\sim}~\int_V \mathrm{d}^n x ~ J^{\mu}~  d_{\mu}\epsilon 
~\sim~-\int_V \mathrm{d}^n x ~ \epsilon~  d_{\mu}   J^{\mu} $$

for arbitrary $x$-dependent infinitesimal $\epsilon(x)$. Equation (11) is precisely OP's sought-for eq. (*).

VI) Equation (11) implies (via the [fundamental lemma of calculus of variations](http://en.wikipedia.org/wiki/Fundamental_lemma_of_calculus_of_variations)) the conservation law 

$$\tag{12} d_{\mu}J^{\mu}~\approx~0, $$ 

in agreement with Noether's theorem.

References:

1. P.K. Townsend, _Noether theorems and higher derivatives,_ [arXiv:1605.07128](http://arxiv.org/abs/1605.07128).

2. G. Barnich, F. Brandt and  M. Henneaux, _Local BRST cohomology in gauge theories,_ Phys. Rep. 338 (2000) 439, [arXiv:hep-th/0002245](http://arxiv.org/abs/hep-th/0002245).  

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$^1$ Since the $x$-dependence of $\epsilon(x)$ is supposed to be just an artificial trick imposed by us, we may assume that there do not appear any derivatives of $\epsilon(x)$ in the transformation law (3), as such terms would vanish anyway when $\epsilon$ is $x$-independent.

$^2$ _Notation:_ The $\sim$ symbol means equality modulo boundary terms. The $\approx$ symbol means equality modulo eqs. of motion.

$^3$ A _quasisymmetry_ of a local action $S=\int_V d^dx ~{\cal L}$ means that the infinitesimal change $\delta S\sim 0$ is a boundary term under the quasisymmetry transformation.