Is the inequality of the random matrices correct?

It's not correct.

Take random variable $xi_i$ as $mathbb{P}(xi_i=1)=1$ for each $i$. Then $Var[xi_i]=0$ for each $i$ and thus $sigma = 0$. l.h.s. hold with probability $1$ while r.h.s is smaller than $1$ is take $t$ sufficiently large.

So, it's a corollary of their Theorem 1.5. I'll restate that inequality and work it into the form you have.

To avoid a conflict of notation, I'm going to change the notation in the paper you linked such that $tmapstoalpha$ and $sigmamapstosigma_0$, and I will hold your notation fixed. Also, that theorem is stated for the rectangular case where $mathbf{B}_k$ has dimension $d_1times d_2$. Here we have $d_1=d_2=m$, and I've simplified my restatement of their theorem below to match the square case.

In that notation, the inequality in Theorem 1.2. reads

$$Pbigg( bigg| sum_k xi_k mathbf{B}_k bigg| geq alphabigg) leq 2m cdot e^{-alpha^2/2sigma_0^2},$$ where $xi_k$ are either independent standard Gaussian or independent Rademacher (which is the same as random signs as you have), and where $sigma_0^2=left| sum_k mathbf{B}_k^2 right|$ (that's the square version of their definition of $sigma_0$).

Now it's just a game of translating this equality into yours. First, let $t=alpha/sigma_0$. Then the inequality becomes $$Pbigg( bigg| sum_k xi_k mathbf{B}_k bigg| geq tsigma_0bigg) leq 2m cdot e^{-t^2/2}.$$ That's a first step.

Now let's deal with the $mathbb{E}xi$ sum. Clearly $mathbb{E}xi=0$, since $xi$ takes values $pm1$ with equal probability. So, the first sum $sum_{i=1}^n mathbb{E}[xi]A_i$ is identically 0. Thus it does not affect the inequality at all. Even better, the variance of the Rademacher variates is $0.5cdot 1^2 + 0.5cdot (-1)^2=1$, so that $sigma=sigma_0$.

So, the inequality you have is exactly the same as the one here, with $A$ changed to $mathbf{B}$ and some extra terms that have no effect. Let me know if any details need clarification and I'll edit those in.