I agree with @Trurl's comment on this. Decision theory is more a component of optimal control theory rather than game theory. In control theory you have an environment you are navigating through to maximize some utility, and the decision theory component is in how to design your navigation as best as possible (often due to information and computational constraints). Game theory is about maximizing your utility in a competition against another player who is trying to do the same. It's possible to define your "environment" in control theory as playing against another player, but this leads to inhomogeneities in the environment most people working in optimal control don't normally deal with. Likewise, a game theorist could phrase an optimal control problem as a "game against the environment", but this would lead to payoff matricies that change drastically over time and with large branching factors.
Your reasoning is correct, the player can deviate to any strategy. Thus, your suggested strategies and randomization device do not constitute a correlated equilibrium.
The way you try to construct it, the correlated equilibrium cannot get more than Nash equilibrium. The main point is that you observe your recommendation but not the recommendation to the other player. However, you construct it in such a way that the players should play either $(T,L)$ or $(B,R)$. Thus, if the a player is assigned a strategy, she knows the assigned strategy of the other player. Let me illustrate this point by constructing a correlated equilibrium in your example.
Consider the following recommendations. With probability $x_1=\frac{1}{4}$ the players are assigned the strategies $(T,L)$, with probability $x_2=\frac{3}{8}$ the strategies $(M,L)$, and with probability $x_3=\frac{3}{8}$ the strategies $(M,R)$.
Now, if Player 1 observes the recommendation $T$, she is sure that player 2 was recommended $L$ and $T$ is a best reply. If player one observes the recommendation $M$, she is not sure whether player 2 is supposed to play $L$ or $R$. However, she knows that both strategies are equally likely given her own recommendation. Thus, she is indifferent between playing either of her strategies. In particular, $M$ is a best reply.
If Player 2 observes the recommendation $L$, she knows that Player 1 is recommended $T$ with probability $\frac{2}{5}$ and $M$ with probability $\frac{3}{5}$ (Bayesian updating). In this case, she is indifferent between playing $L$ or $C$, so $L$ is a best reply. If player 2 observes the recommendation $R$, she knows that player 1 is recommended $M$ and $R$ is a best reply. So overall you have a correlated equilibrium.
Best Answer
The following is an opinion regarding your first question ("has game theory any practical relevance?") by Ariel Rubinstein, a famous game theorist. I do not agree with him, but he has an interesting argument :
From an article in the Frankfurter Allgemeine (hope this is not infringing on anyone's copyright...)
I think Rubinstein develops these ideas in yet another paper or interview but I could not find it (in this other paper, he notably argued that if game theory has ever been useful to anyone, it is to the wealthy and to the powerfull and that, as a consequence, it did not help foster anything like "social justice"). Anyone has a clue on that?