Let $S$ denote the Schwartz space of functions $\mathbb R\to\mathbb R$ with its usual topology. Let $S_e$ and $S_o$ denote the subspaces consisting of even and odd functions, respectively. Suppose I have a subspace $X\subset S$ (just a linear subspace without any topological assumptions) so that for any $f\in S$
$$ f\in S_o \iff \left(\int fg=0\text{ for all }g\in X\right) . $$
Does it follow that $X$ is dense in $S_e$? It is not hard to check that $X\subset S_e$ and I feel it should be dense, but I don't see how to prove it. If I replace $X$ with its closure, I get an alternative formulation of the question: "Assume additionally that $X$ is closed. Does it follow that $X=S_e$?"
The density result is true if I replace $S$ with $L^2$, but $S$ is not complete under the $L^2$ inner product. Assuming the result is false, there is a function $h\in S_e\setminus\bar X$, and one could then project to the orthogonal complement of $\bar X$ to produce a nonzero function $h'\in S_e$ so that $\int h'g=0$ for all $g\in\bar X$, a contradiction. If one attacks the problem this way, the problem is in showing that the $L^2$-orthogonal projection from $S_e$ to $\bar X$ is well defined (actually maps Schwartz functions to Schwartz functions). The closures I have taken are in the natural topology of $S$, not $L^2$ topology, so it's not trivial to make sense of the projection.
Consider e.g. $X = \{g \in S_e: g(1) = 0\}$. This is not dense in $S_e$, but I claim it satisfies your condition.
Note that if $f \in S \backslash S_o$, $f(x) + f(-x)$ is nonzero on a nonempty open subset $A$ of $(0,\infty)$. Wlog it is positive on $A$. We can then take $g_1 \in S$ supported in a closed interval $J$ contained in $A \backslash \{1\}$ and positive there. Then $g_1(x) + g_1(-x) \in X$ and $$\int_R (g_1(x) + g_1(-x)) f(x)\; dx = \int_J g_1(x) (f(x) + f(-x))\; dx > 0$$