lift an isomorphism of quotients to isomorphism 
Let $R$ be a local ring with maximal ideal $M$ and $I$ an ideal with $IM=(0)$. Let $A$ and $B$ be two $R$-algebras and $B$ flat over $R$. Let $\phi: A\to B$ be an $R$-morphism such that the induced homomorphism $\bar{\phi}: A/IA\to B /IB$ is an isomorphism. I want to show that $\phi$ is an isomorphism.

If $A,B$ were finite generated (e.g. noetherian) then the flat $R$-module would be free and therefore projective. this would imply injectivity. Lemma of Nakayama applied to cokernel of $\phi$ would imply injectivity. but neither the result "flat + fin. generated $\Rightarrow$ free", nor Nakayama lemma are available, because we not assumed that $A,B$ are finitely generated.
Surjectivity I think works as follows: Since $\bar{\phi}$ isomorphism, we find $b_1,...,b_k \in B$ and $l_1,..., l_n$ with $1+\sum_{j=1}^n b_j \cdot l_j \in \phi(A)$. $phi$ is a homomorphism of algebras and thus $(1+\sum_{j=1}^n b_j \cdot l_j)^2= 1+ 2 \sum_{j=1}^n b_j \cdot l_j \in \phi(A)$ (we used $I \subset M$, thus $I^2= I M =0$). this implies that $\sum_{j=1}^n b_j \cdot l_j \in \phi(A)$ and therefore $1_B \in \phi(A)$. this is surjectivity. Does it make sense? 
I don't know how to prove injectivity.
 A: I think we can generalize to the following claim:

Claim: Let $R$ be a ring with an ideal $I$ such that $I^\ell = 0$. Let $A,B$ be two $R$-modules and let $\phi : A \to B$ be an $R$-module homomorphism such that $\overline{\phi} : A/IA \to B/IB$ is an isomorphism. If $B$ is flat, then $\phi$ is an isomorphism.

Here is a lemma:

Lemma: Let $R$ is a ring with an ideal $I$ such that $I^\ell = 0$. If $A$ is an $R$-module with $A = IA$ then $A = 0$.

Proof of Claim: We show surjectivity first. It basically follows from (9) of Tag 00DV but I write out some details here. The condition that $\overline{\phi}$ is surjective is equivalent to $B/IB = (IB + \phi(A))/IB$; thus $B = IB + \phi(A)$; thus $B/\phi(A) = I(B/\phi(A))$; thus $B/\phi(A) = 0$ by the Lemma.
For injectivity, set $K := \ker \phi$ so that we have an exact sequence $0 \to K \to A \to B \to 0$ of $R$-modules; since $B$ is flat, this sequence stays exact when we tensor with $R/I$ (see Tag 00HL); thus $K/IK = 0$; thus $K = 0$ by the Lemma.
A: I think I have almost solved it myself:
surjective: $\bar{\phi}$ is bijective and thus for every $b \in B$ there exist a $a \in A, l_i \in I, b_i \in B$  with $\phi(a)= b + \sum_i ^m l_i b_i $. $\phi$ is a $R$-homomorphism and therefore for every $l \in I$ $\phi(i \cdot a)= l \cdot \phi(a)= l \cdot b$. we deduce $I \cdot B \subset \phi(A)$ and since $b$ was arbitrary, $\phi$ is surjective. 
injective: $B$ flat implies torsion-free. let $a \in ker(\phi)$. since $\bar{\phi}$ isomorphism, $a \in I \cdot A$, $a= \sum_i ^n l_i a_i$ with $l_i \in I, a_i \in A \setminus A \cdot I$. 
I solved only the case $n=1$, i.e. $a= l \cdot \tilde{a}, l \in I, \tilde{a} \in A \setminus A \cdot I$. in this case $\bar{\phi}(\tilde{a}) \neq 0$ and thus $\phi(\tilde{a}) \neq 0$. by assumption $\phi(a)= l \cdot \phi(\tilde{a}) =0$ and therefore $\phi(\tilde{a}) \in B$ is torsion element. contradiction.
only unsolved case is $n >1$.
