I interpret your game as follows. $A$ chooses $\alpha\in[0,1]$, $B$ chooses some measurable $\sigma:X\to[0,1]$, resulting in a probability $$u(\alpha,\sigma)=\int [\alpha(1-\sigma)\ d\mu + (1-\alpha)\sigma\ d\nu]$$ of $B$ guessing incorrectly. Here, the players move simultaneously, $A$ wants to maximize $u$ and $B$ wants to minimize $u$.

It's possible to explicitly compute a value for this game in general, but let's consider a special case (which is, in a sense, not very special) that cleans up notation, in order to specifically address the question of how the value of the game relates to total variation.

Specifically, for a given nondecreasing $f:[0,1]\to[0,2]$, suppose $X=[0,1]$, measure $\mu$ has Lebesgue density $f$, and $\nu$ has Lebesgue density $2-f$. Then, consider $\alpha^*:=1-\tfrac12 f(\tfrac12)$ and $\sigma^*:=\mathbf1_{[\tfrac12,1]}$. One can verify that this is a Nash equilibrium (a.k.a. saddle point): $\alpha^*$ maximizes $u(\cdot,\sigma^*)$ and $\sigma^*$ minimizes $u(\alpha^*,\cdot)$, and the value it produces is exactly $$v = \int_0^{\tfrac12}f.$$ Meanwhile, the total variation between $\mu$ and $\nu$ can be computed as $$TV = 2\int_0^1 \max\{f-1, 0\}.$$

So the question, restricted to this special case, reduces to: When looking at nondecreasing functions $f:[0,1]\to[0,2]$, can one express $\int_0^{\tfrac12}f$ as a function of $2\int_0^1 \max\{f-1, 0\}$?

If it so happens that $f(\tfrac12)=1$, then $$\tfrac12 TV=\int_0^1 \max\{f-1, 0\}=\int_{\tfrac12}^1 (f-1)=\left(\int_0^1 f \right) - \tfrac12 - \int_0^{\tfrac12} f = \tfrac12-v,$$ implying $v=\tfrac{1-TV}2$.

But, if $\int_0^1 \max\{f-1, 0\}\neq\int_{\tfrac12}^1 (f-1)$, then the above algebra shows that $v\neq\tfrac{1-TV}2$. Finally, it's easy to find two continuous, strictly increasing functions $f,\hat f:[0,1]\to[0,2]$ such that $\int_0^1 \max\{f-1, 0\}=\int_0^1 \max\{\hat f-1, 0\}$ but $f(\tfrac12)=1\neq \hat f(\tfrac12)$.

So in summary, the value cannot generally be expressed as a function of total variation. However, the original post reflected a good intuition. If we restrict to the case that there is "enough symmetry" between $\mu$ and $\nu$ that player $A$ could choose a 50/50 mixture in equilibrium, then the value is a function of total variation.

I interpret your game as follows. $A$ chooses $\alpha\in[0,1]$, $B$ chooses some measurable $\sigma:X\to[0,1]$, resulting in a probability $$u(\alpha,\sigma)=\int [\alpha(1-\sigma)\ d\mu + (1-\alpha)\sigma\ d\nu]$$ of $B$ guessing incorrectly. Here, the players move simultaneously, $A$ wants to maximize $u$ and $B$ wants to minimize $u$.

It's possible to explicitly compute a value for this game in general, but let's consider a special case (which is, in a sense, not very special) that cleans up notation, in order to specifically address the question of how the value of the game relates to total variation.

Specifically, for a given nondecreasing $f:[0,1]\to[0,2]$ with $\int_0^1 f=1$, suppose $X=[0,1]$, measure $\mu$ has Lebesgue density $f$, and $\nu$ has Lebesgue density $2-f$. Then, consider $\alpha^*:=1-\tfrac12 f(\tfrac12)$ and $\sigma^*:=\mathbf1_{[\tfrac12,1]}$. One can verify that this is a Nash equilibrium (a.k.a. saddle point): $\alpha^*$ maximizes $u(\cdot,\sigma^*)$ and $\sigma^*$ minimizes $u(\alpha^*,\cdot)$, and the value it produces is exactly $$v = \int_0^{\tfrac12}f.$$ Meanwhile, the total variation between $\mu$ and $\nu$ can be computed as $$TV = 2\int_0^1 \max\{f-1, 0\}.$$

So the question, restricted to this special case, reduces to: When looking at nondecreasing functions $f:[0,1]\to[0,2]$, can one express $\int_0^{\tfrac12}f$ as a function of $2\int_0^1 \max\{f-1, 0\}$?

If it so happens that $f(\tfrac12)=1$, then $$\tfrac12 TV=\int_0^1 \max\{f-1, 0\}=\int_{\tfrac12}^1 (f-1)=\left(\int_0^1 f \right) - \tfrac12 - \int_0^{\tfrac12} f = \tfrac12-v,$$ implying $v=\tfrac{1-TV}2$.

But, if $\int_0^1 \max\{f-1, 0\}\neq\int_{\tfrac12}^1 (f-1)$, then the above algebra shows that $v\neq\tfrac{1-TV}2$. Finally, it's easy to find two continuous, strictly increasing functions $f,\hat f:[0,1]\to[0,2]$ such that $\int_0^1 \max\{f-1, 0\}=\int_0^1 \max\{\hat f-1, 0\}$ but $f(\tfrac12)=1\neq \hat f(\tfrac12)$.

So in summary, the value cannot generally be expressed as a function of total variation. However, the original post reflected a good intuition. If we restrict to the case that there is "enough symmetry" between $\mu$ and $\nu$ that player $A$ could choose a 50/50 mixture in equilibrium, then the value is a function of total variation.