Exercise 9.5.21 in Grillet (Jacobson radical, Nakayama's Lemma) I've managed to prove the following exercise:

(21.) Let $\mathfrak{m}$ be a maximal ideal of a commutative ring $R$. Prove the following: if $A$ is a finitely generated $R$-module, and $x_1, x_2, \ldots, x_n$ is a minimal generating subset of $A$, then $x_1 + \mathfrak{m} A, \ldots, x_n + \mathfrak{m} A$ is a basis of $A/\mathfrak{m}A$ over $R / \mathfrak{m}$.

in Grillet's Abstract Algebra, page 377, only under the hypothesis that $\mathfrak{m}$ is the only maximal ideal, i.e., that $R$ is a local ring.
Weak proof: Firstly, the scalar multiplication is defined as $(r\!+\!\mathfrak{m})(a\!+\!\mathfrak{m}A) \!:=\! ra\!+\!\mathfrak{m}A$. To see this is well defined, notice that if $r\!+\!\mathfrak{m} \!=\! 0$, i.e. $r\!\in\!\mathfrak{m}$, then $ra\!\in\!\mathfrak{m}A$, i.e. $ra\!+\!\mathfrak{m}A\!=\!0$; and also that if $a\!+\!\mathfrak{m}A\!=\!0$, i.e. $a\!=\!ma'\!\in\!\mathfrak{m}A$, then $ra\!=\!rma'\!\in\!\mathfrak{m}A$, i.e. $ra\!+\!\mathfrak{m}A\!=\!0$. Actually, the $R/\mathfrak{m}$-module $A/\mathfrak{m}A$ is constructed by first creating the $R$-module $A/\mathfrak{m}A$ (this is possible because $\mathfrak{m}A$ is a submodule of $A$, since $\mathfrak{m}$ is an ideal of $R$) and then turning it into an $R/\mathfrak{m}$-module (this is possible because $\mathrm{Ann}_R(A/\mathfrak{m}A)\supseteq\mathfrak{m}$).
To prove $x_1\!+\!\mathfrak{m}A,\ldots,x_n\!+\!\mathfrak{m}A$ generate $A/\mathfrak{m}A$, we must show that $(R/\mathfrak{m})(x_1\!+\!\mathfrak{m}A)\!+\!\ldots\!+\!(R/\mathfrak{m})(x_n\!+\!\mathfrak{m}A)\!=\!A/\mathfrak{m}A$, which means $Rx_1\!+\!\ldots\!+\!Rx_n\!+\!\mathfrak{m}A\!=\!A/\mathfrak{m}A$, or equivalently, $Rx_1\!+\!\ldots\!+\!Rx_n\!+\!\mathfrak{m}A\!=\!A$, but this is true since $Rx_1\!+\!\ldots\!+\!Rx_n\!=\!A$ by hypothesis. Alternatively, we could argue that since $x_1,\ldots,x_n$ generate the $R$-module $A$, they generate the $R$-module $A/\mathfrak{m}A$, and therefore generate the $R/\mathfrak{m}$-module $A/\mathfrak{m}A$.
If $x_1\!+\!\mathfrak{m}A,\ldots,x_n\!+\!\mathfrak{m}A$ were $R/\mathfrak{m}$-linearly dependent, then WLOG $x_n\!+\!\mathfrak{m}A$ could be expressed as a $R/\mathfrak{m}$-linear combination of the others, so already $x_1\!+\!\mathfrak{m}A,\ldots,x_{n-1}\!+\!\mathfrak{m}A$ would generate $A/\mathfrak{m}A$, which would mean $Rx_1\!+\!\ldots\!+\!Rx_{n-1}\!+\!\mathfrak{m}A\!=\!A$. But since $J(R)\!=\!\mathfrak{m}$, by Nakayama's Lemma this would mean $Rx_1\!+\!\ldots\!+\!Rx_{n-1}\!=\!A$, a contradiction with the hypothesis on minimality of  $x_1,\ldots,x_n$. $\blacksquare$
Question: how can I prove the general version of the exercise? I am somewhat skeptical of the claim...
 A: Using Dylan's comment, we get the following counterexample. Let $K$ and $L$ be fields, put
$$
R=K\times L,\quad A=\mathfrak m=K,\quad n=1,\quad x_1=(1,0),
$$
and observe
$$
A/\mathfrak mA=K/KK=K/K=0.
$$
A: You are quite right to be skeptical of the claim, Leon, because I think it is false. Here is a counterexample.
Let $k$ be a field,  $R$ be the ring $R=k[X,Y]/\langle Y^2-X^3\rangle=:k[x,y]$ and  $A$ the ideal $A=\langle x,y\rangle\subset R$.
The chosen set of generators $x_1=x,x_2=y$  of the ideal $A$ is already   minimal  since $A$ is not principal.
Now choose the maximal ideal ${\frak m}=\langle x-1,y-1 \rangle$.
We have $R/{\frak m}=k$ and ${\frak m}A=\langle x-1,y-1 \rangle \langle x,y \rangle= \langle  x^2-x,xy-y,yx-x,y^2-y\rangle  $ so that
$A/{\frak m}A=\langle x,y\rangle/\langle  x^2-x,xy-y,yx-x,y^2-y\rangle=
\langle \bar x \rangle$, a one-dimensional $k$ vector space.
But then the two classes $x+{\frak m}A ,\; y+{\frak m}A$ can obviously not be a basis of this one-dimensional vector space.   
Edit
At Leon's request I'll explain in detail why the ideal $A\subset R$ is not principal (this should  be skipped by whoever is familiar with the underlying algebraic geometry of the situation).       
Suppose $A=\langle f(x,y)\rangle $ for some  $ f(X,Y)\in k[X,Y]$ with $f(X,Y)=0$. Since $f\in A$ and since $y^2=x^3$, we can write $f(x,y)=xg(x)+yh(x)$ with $g(X),h(X)\in k[X,Y]$ .    
i) Since $x\in A=\langle f(x,y) \rangle$ we must have $x=f(x,y) a(x,y)$ for some $a(X,Y)\in k[X,Y] \:$ . Lifting to actual polynomials we get  $X=(Xg(X)+Yh(X))a(X,Y)+(Y^2-X^3)P(X,Y)$.
Looking at linear terms, we see that this is only possible if $g(0)=:c\neq0$ and $h(0)=0$ so that $$g(X)=cX+X^2.l(X) ,\quad  h(X)=Xm(X) \quad (*)$$  
ii) But then $y\in A=\langle f(x,y) \rangle$ similarly translates into 
$Y=(Xg(X)+YhX))b(X,Y)+(Y^2-X^3)Q(X,Y)$ and so, using     $(*)$ ,
$Y=[X.(cX+X^2l(X))+YXm(X)].b(X,Y)+(Y^2-X^3)Q(X,Y)$
But this is impossible because the right-hand side has zero as its  linear term.
We have proved that the assumption that $A$ is principal is contradictory.
