Does an asymmetric Bernoulli matrix satisfy the RIP?
As others have stated in the comments, the answer is "No".  The non-zero mean of the matrix dictates that a nonzero mean vector (say, all ones), will have substantially higher gain than a random vector with zero mean (say uniformly random +1,-1).

Consider the squared norm of A times a constant vector y is expected to be n*(p*N)^2.  (iteration of expectations)

The squared norm of A times a vector x drawn uniformly from (-1,+1) is expected to be n*(p*N).  (calculable by sum of variances of Binomial distribution)

The norms of x and y are the same, but the expectation of transformed norms differ by a factor of p*N -- diverging as the dimensions grow large.  

Here's matlab code to help demonstrate.

n=2000;
N=1000;
p=.9;
A=double(rand(n,N)&lt;p); 
x=sign(randn(N,1)); 
y=ones(N,1);
Ex_normSqAx = n*(N*p);  % E[ squared norm of A times random signs ]
Ex_normSqAy = n*(N*p)^2; % E[ squared norm of A times constant vector ]
normSqAx = norm(A*x)^2;
normSqAy = norm(A*y)^2;