An extension of the noise operator
I'll answer the second part of the question.

I. Eigenvalues and Eigenfunctions

Let's first consider the one dimensional case $n=1$. It is easy to check that the operator $R_{p_1,p_2}$ has two eigenfunctions: $1$ and 
$$\xi(x) = (p_1+p_2)x - p_1 =
\begin{cases}
-p_1, &amp;\text{ if } x =0,\\
p_2, &amp;\text{ if } x =1.
\end{cases}$$
with eigenvalues $1$ and $1-p_1 - p_2$, respectively.

Now consider the general case. For $S\subset \{1,\dots,n\}$, let $\xi_S(x) = \prod_{i\in S} \xi(x_i)$. Observe that $\xi_S$ is an eigenfunction of $R_{p_1,p_2}$. Indeed since all variables $x_i$ are independent, we have
\begin{align*}
R_{p_1,p_2}(\xi(x)) &amp;= R_{p_1,p_2}\left(\prod_{i\in S} \xi(x_i)\right) = \prod_{i\in S} R_{p_1,p_2}(\xi(x_i)) \\
&amp;= \prod_{i\in S}\left((1-p_1 - p_2)\xi(x_i)\right) = (1-p_1 - p_2)^{|S|} \xi_S(x).
\end{align*}

We get that $\xi_S(x)$ is an eigenfunction of $R_{p_1,p_2}$ with eigenvalue $(1-p_1-p_2)^{|S|}$ for every $S\subset \{1,\dots,n\}$. Since functions $\xi_S(x)$ span the whole space, $R_{p_1,p_2}$ has no other eigenfunctions (that are not linear combinations of $\xi_S(x)$).

II. Multiplicative Property

In general, the “multiplicative property” doesn't hold for $R_{p_1,p_2}$ since the eigenbasis of $R_{p_1,p_2}$ depends on $p_1$ and $p_2$. However, we have 
$$R_{p_1,p_2}^2 = R_{p_1',p_2'},$$
where $p_1' = 2p_1- (p_1+p_2)p_1$ and $p_2' = 2p_2- (p_1+p_2)p_2$. To verify that, first note that $R_{p_1,p_2}$ and $R_{p_1',p_2'}$ have the same set of eigenfunctions $\{\xi_S\}$. We have,
$$R_{p_1,p_2}^2(\xi_S) = (1-p_1-p_2)^{2|S|} \xi_S=(1-p_1'-p_2')^{|S|}\xi_S = R_{p_1',p_2'} (\xi_S) $$
since
\begin{align*}
 1-p_1'-p_2' &amp;= 1 - p_1\cdot (2- (p_1+p_2)) - p_2\cdot (2- (p_1+p_2)) \\
&amp;= 1 - (p_1+p+2) (2- (p_1+p_2)) \\
&amp;= 1 - 2(p_1 +p_2) + (p_1 + p_2)^2 = (1-p_1-p_2)^2.
\end{align*}

III. Relation to the Bonami—Beckner operator

Let us think of functions from $\{0,1\}^n$ to ${\mathbb R}$ as polylinear polynomials. Let $\delta = \frac{1}{2}\cdot \frac{p_1-p_2}{p_1+p_2}$. Consider the operator 
$$A_{\delta}(f) = f(x_1 + \delta, \dots, x_n + \delta).$$ 
It maps every multilinear polynomial $f$ to a multilinear polynomial $A[f]$. We have,
$$R_{p_1,p_2}(f) = A_{\delta}^{-1} T_{\varepsilon}A_{\delta}(f),$$
where $\varepsilon = 1 - p_1 - p_2$. Note that parts I and II follow from this formula and properties of the Bonami—Beckner operator.