First, we point out that $c \ge \cot \frac{\pi}{4022}$. Consider the polynomial $P(x) = x^{2012} - x$. Changing the sign of coefficients of $P(x)$, we obtain four polynomials $P(x)$, $-P(x)$, $Q(x) = x^{2012} + x$ and $-Q(x)$. Note that $P(x)$ and $-P(x)$ have the same roots; one of the roots is $z_1 = \cos \frac{1006}{2011}\pi + i \sin \frac{1006}{2011}\pi$. $Q(x)$ and $-Q(x)$ have the same roots, and are the opposite number of the roots of $P(x)$. Thus $Q(x)$ has a root $z_2 = -z_1$. Then,
$$c \ge \min\left(\frac{|\operatorname{Im} z_1|}{|\operatorname{Re} z_1|}, \frac{|\operatorname{Im} z_2|}{|\operatorname{Re} z_2|}\right) = \cot \frac{\pi}{4022}.$$
Next, we show that the answer is $c = \cot \frac{\pi}{4022}$. For any
$$P(x) = x^{2012} + a_{2011}x^{2011} + a_{2010}x^{2010} + \dots + a_1x + a_0,$$
we obtain a polynomial
$$R(x) = b_{2012}x^{2012} + b_{2011}x^{2011} + b_{2010}x^{2010} + \dots + b_1x + b_0,$$
by changing sign of some coefficient of $P(x)$, where $b_{2012} = 1$, and for $j = 1, 2, \dots, 2011$,
$$b_j = \begin{cases} |a_j|, & j \equiv 0, 1 \pmod 4, \\ -|a_j|, & j \equiv 2, 3 \pmod 4. \end{cases}$$
We show that, for each root $z$ of $R(x)$, we have $|\operatorname{Im} z| \le c |\operatorname{Re} z|.$
We prove this result by contradiction. Suppose there is a root $z_0$ of $R(x)$, such that $|\operatorname{Im} z_0| > c |\operatorname{Re} z_0|$, then $z_0 \ne 0$ and either the angle of $z_0$ and $i$ is less than $\theta = \frac{\pi}{4022}$, or the angle of $z_0$ and $-i$ is less than $\theta$. Suppose that the angle of $z_0$ and $i$ is less than $\theta$; for the other case, we need only consider the conjugate of $z_0$. There are two cases:

If $z_0$ is on the first quadrant (or imaginary axis), suppose that $\angle(z_0, i) = \alpha < \theta$, where $\angle(z_0, i)$ is the least angle that rotates $z_0$ to $i$ anticlockwise. For $0 \le j \le 2012$, if $j \equiv 0, 2 \pmod 4$, then $\angle(b_j z_0^j, 1) = j\alpha \le 2012\alpha < 2012\theta$.
If $j \equiv 1, 3 \pmod 4$, then $\angle(b_j z_0^j, i) = j\alpha < 2011\theta$ and $\angle(b_1 z_0, i) = \alpha$. Thus, the principal argument of $b_j z_0^j \in [2\pi - 2012\alpha, 2\pi) \cup [0, \frac{1}{2}\pi - \alpha]$. The vertex angle of this angle-domain is $2012\alpha + \frac{1}{2}\pi - \alpha = \frac{1}{2}\pi + 2011\alpha < \pi$. And $b_j z_0^j, 0 \le j \le 2012$, are not all zero, so their sum cannot be zero.

If $z_0$ is at the second quadrant, suppose that $\angle(i, z_0) = \alpha < \theta$, if $j \equiv 0, 2 \pmod 4$, then $\angle(1, b_j z_0^j) = j\alpha < 2012\theta$. If $j \equiv 1, 3 \pmod 4$, then $\angle(i, b_j z_0^j) = j\alpha \le 2011\alpha < \frac{\pi}{2}$. Thus, every principle argument of $b_j z_0^j \in [0, \frac{\pi}{2} + 2011\alpha]$. Since $\frac{\pi}{2} + 2011\alpha < \pi$, and $b_j z_0^j, 0 \le j \le 2012$, are not all zero, so their sum cannot be zero.

Summing up, the least real number $c = \cot \frac{\pi}{4022}$. $\square$