Uncorrelated model error
If $\mathbf{w_k} $ was temporally correlated, it would have memory and would be equivalent to an augmented state.  In other words, it can be written as a state variable with its own independent noise term. You would be able to rewrite the equation to the same form you originally showed.  It is the most basic form of the model.  

The symmetry of $Q_k$ is because $E\{ ab\}=E\{ba\}$ for any terms $a$ and $b$ of the vector $\mathbf{w_k}$

Edit to address comments.

Let $\mathbf{x}=\mid x_0,x_1, \dots x_{N-1} \mid^T$ be a random Gaussian distributed random vector. 
$$
E\left\{ \mathbf{x} \mathbf{x}^T \right\}=\mathbf{R} \quad \text{Is typically Full rank and also typically} = \sigma^2 \mathbf{I}
$$
when the elements of $\mathbf{x}$ are independent.
There are some exceptions to full rank, like like a linear dependence of some of the elements of $\mathbf{x }$ such as $x_i=x_j$ of more generally $x_i=\sum_{k=0, k\ne i}^{N-1} \alpha_k x_k$ .

The easiest example is to consider $E\{ x_i x_i \} = E\{ x_i^2\}$  .  A Gaussian random variable can have zero mean but a variance $\sigma^2$. The matrix $\mathbf{R}$ would have nonzero diagonal terms, and is Full Rank.  

The Matrix $\mathbf{R}$ isn't required to be diagonal and has nonzero off diagonal terms when $E\{x_i x_j\} \ne 0$ for some $i$ and $j$.  For your problem, your book will likely show the covariance of the evolution of the the state vector and that will most likely have off diagonal nonzero elements. 
$$
E\left\{ \mathbf{x}\right\} E\left\{ \mathbf{x}^T \right\} \quad \text{Is rank one} 
$$