This isn't an answer to your question because I have no idea whether this argument can be carried out in $RCA_0$, but this fact doesn't appear to be mentioned anywhere easily accessible through googling about Monsky's theorem so it seems good to document it somewhere (assuming I didn't make a mistake!). We can say the following (unless I am horribly mistaken about something!):

Claim 1: A system of polynomial equations $P_1(x_1, \dots x_n) = \dots = P_k(x_1, \dots x_n)$ over $\mathbb{Q}$ has a solution over $\mathbb{R}$ iff it has a solution over the real algebraic numbers $\mathbb{R} \cap \overline{\mathbb{Q}}$.

Claim 2: For a fixed positive integer $k$, Monsky's theorem for subdivisions into $2k+1$ triangles is equivalent to the statement that no member of a specific finite set of systems of polynomial equations has a solution over $\mathbb{R}$.

The upshot is that Monsky's theorem is true over $\mathbb{R}$ iff it's true over $\mathbb{R} \cap \overline{\mathbb{Q}}$, and hence that we only need to extend the $2$-adic valuation to the finite extension of $\mathbb{Q}$ generated by the coordinates of a potential algebraic counterexample, which is just a number field $K$. And this can be done with no shenanigans whatsoever, since it suffices to choose a prime ideal of $\mathcal{O}_K$ lying over $(2)$. So there is no need, as far as I can tell, to consider even a single transcendental number in the argument.

Claim 1 follows from the fact that $\mathbb{R}$ is elementary equivalent to any real closed field but I believe it should also have a more elementary proof using elimination theory; hopefully someone else can fill in the details and say what exactly is needed in the proof.

Claim 2 is the usual reduction of Euclidean geometry to the theory of real closed fields: an equal-area dissection of the unit square into $2k+1$ triangles consists of the coordinates of a finite collection of points together with one of a finite set of finitely many assertions that these points 1) lie inside the unit square, and 2) define triangles with area $\frac{1}{2k+1}$ (we may want to draw different lines between the points to define different sets of triangles but for $k$ fixed there are only finitely many possible configurations). A real number $x$ lies in $[0, 1]$ iff $\exists y_1 : x = y_1^2$ and $\exists y_2 : 1 - x = y_2^2$, and the area of a triangle is a quadratic polynomial in the coordinates of its vertices, so the conclusion is that for fixed $k$ the existence of an equal-area dissection into $2k+1$ triangles is equivalent to the existence of a solution to one of a finite set of systems of quadratic polynomials.

This raises the funny possibility that for each fixed $k$ it might be possible to prove Monsky's theorem for dissections into $2k+1$ triangles in some suitably weak theory, but it might not be possible to prove the theorem quantified over all $k$. (At least I think?)

This isn't an answer to your question because I have no idea whether this argument can be carried out in $RCA_0$, but this fact doesn't appear to be mentioned anywhere easily accessible through googling about Monsky's theorem so it seems good to document it somewhere (assuming I didn't make a mistake!). We can say the following (unless I am horribly mistaken about something!):

Claim 1: A system of polynomial equations $P_1(x_1, \dots x_n) = \dots = P_k(x_1, \dots x_n)$ over $\mathbb{Q}$ has a solution over $\mathbb{R}$ iff it has a solution over the real algebraic numbers $\mathbb{R} \cap \overline{\mathbb{Q}}$.

Claim 2: For a fixed positive integer $k$, Monsky's theorem for subdivisions into $2k+1$ triangles is equivalent to the statement that no member of a specific finite set of systems of polynomial equations has a solution over $\mathbb{R}$.

The upshot is that Monsky's theorem is true over $\mathbb{R}$ iff it's true over $\mathbb{R} \cap \overline{\mathbb{Q}}$, and hence that we only need to extend the $2$-adic valuation to the finite extension of $\mathbb{Q}$ generated by the coordinates of a potential algebraic counterexample, which is just a number field $K$. And this can be done with no shenanigans whatsoever, since it suffices to choose a prime ideal of $\mathcal{O}_K$ lying over $(2)$. So there is no need, as far as I can tell, to consider even a single transcendental number in the argument.

Claim 1 follows from the fact that $\mathbb{R}$ is elementary equivalent to any real closed field but I believe it should also have a more elementary proof using elimination theory; hopefully someone else can fill in the details and say what exactly is needed in the proof.

Claim 2 is the usual reduction of Euclidean geometry to the theory of real closed fields: an equal-area dissection of the unit square into $2k+1$ triangles consists of the coordinates of a finite collection of points together with one of a finite set of finitely many assertions that these points 1) lie inside the unit square, and 2) define triangles with area $\frac{1}{2k+1}$ (we may want to draw different lines between the points to define different sets of triangles but for $k$ fixed there are only finitely many possible configurations). A real number $x$ lies in $[0, 1]$ iff $\exists y_1 : x = y_1^2$ and $\exists y_2 : 1 - x = y_2^2$, and the area of a triangle is a quadratic polynomial in the coordinates of its vertices, so the conclusion is that for fixed $k$ the existence of an equal-area dissection into $2k+1$ triangles is equivalent to the existence of a solution to one of a finite set of systems of quadratic polynomials.