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Edit 2: Sorry for being incomprehensible yesterday. I believe the result is true, at least if $R$ is noetherian. Let me assume that $R$ is noetherian, this makes me feel better.

Then $R$ injects into $\prod_p R_p$, where the product is taken over minimal primes of $R$. It is enough to show the statement for $R_p$. Now, $R_p$ is a local ring with a nilpotent maximal ideal.

So assume (as we may) $R$ is local with a nilpotent maximal idela $\mathfrak m$. Consider the exact sequence $0\rightarrow Ker A \rightarrow R^n\rightarrow Im A\rightarrow 0$. Since $Im A$ is free, it is projective, hence the sequence slits. Thus $Ker A$ is projective (since a direct summand of a free). Since $R$ is local, $Ker A$ is free.

Case 1. $\lambda\not\in \mathfrak m$. Then $\lambda$ is a unit in $R$. Thus the intersection of $Ker A$ and $Ker(A-\lambda)$ is zero. Now $A^2=\lambda A$ implies that $R^n= Ker A\oplus Ker(A-\lambda)$, and we are done.

Case 2. $\lambda\in \mathfrak m$. Choose a basis for $Ker A$, $v_1, \ldots, v_{n-k}$ and extend it to a basis of $R^n$, $w_1, \ldots w_{k}$. The images of $Aw_1,.., A w_k$ mod $\mathfrak m$ are linearly independent and contained in $Ker A$ mod $\mathfrak m$. We may assume that $Aw_i = v_i$ mod $\mathfrak m$. Wrt this basis, $A$ looks like a block matrix $(0 B//0 D)$, where the top $k\times k$ square of $B$ is congruent to $1$ mod $\mathfrak m$ ( i hope i got this right), hence this top right corner is invertible. By plugging in the relation $A^2=\lambda A$ and looking at the top right $k\times k$ corner, we get $D=\lambda \times (identity)$, which implies the claim.

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Yes. If $R$ is an integral domain. Just embedd everything inside the quotient field of $R$, and use the result your result over fields (i believe you that it holds).

Yes. If $R$ is reduced and noetherian. By the previous parts the satement holds for $R/p$, where $p$ is a minimal prime ideal. Hence $Tr A$ is equal $\lambda$ mod $p$, for all minimal primes. Since $R$ is reduced the natural $R\rightarrow \prod_p R/p$ is injective, where the product is taken over the minimal primes.

I believe it should fail if $R$ is not reduced, and i guess the example should exist over $R=k[x]/(x^2)$.

Edit: I dont belive the last statement anymore. The image of $A$ is projective, hence we may split it off. The point is that the statement holds when $n=k$, since then $A$ is invertible. The only thing that worries me slightly is that the kernel might not be free, but maybe everything is ok after passing to a covering, where the Kernel is free.

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Yes. If $R$ is an integral domain. Just embedd everything inside the quotient field of $R$, and use the result your result over fields (i believe you that it holds).

Yes. If $R$ is reduced and noetherian. By the previous parts the satement holds for $R/p$, where $p$ is a minimal prime ideal. Hence $Tr A$ is equal $\lambda$ mod $p$, for all minimal primes. Since $R$ is reduced the natural $R\rightarrow \prod_p R/p$ is injective, where the product is taken over the minimal primes.

I believe it should fail if $R$ is not reduced, and i guess the example should exist over $R=k[x]/(x^2)$.