Since $f(x)$ is an even function on $\mathbb{R}$, and $f(x-1)$ is an odd function on $\mathbb{R}$, we have
$f(-x) = f(x)$, and $f(-x-1) = -f(x-1)$, (1)
Therefore, $f(-x-1) = f(x+1)$, (2)
From (1) and (2), we get $f(x+1) = -f(x-1)$ (3) always holds,
Therefore, $f(x-1) = -f(x-3)$ (4)
From (3) and (4), we get $f(x+1) = f(x-3)$ always holds,
Therefore, the period of the function is 4.
Since the graph of $f(x)$ is translated one unit to the right to get an odd function, we have $f(0-1) = 0$, which means $f(-1) = 0$. By the property of even functions, we know $f(1) = 0$, and by periodicity, we know $f(3) = 0$.
Given $f(2) = -1$, we have $f(-2) = -1$, and from $f(x+1) = -f(x-1)$, we know $f(0) = 1$, thus $f(4) = 1$.
Therefore, we have $f(1) + f(2) + f(3) + f(4) = 0$.
Therefore, $f(0) + f(1) + f(2) + f(3) + \ldots + f(2012) = f(0) = 1$.
Hence, the correct answer is $\boxed{\text{B}}$.