I am suspicious of your result. 

The three torus $\mathbb{T}^3$ is well-known to not admit any metric of positive scalar curvature. 

Let $g_0$ be the flat metric on $\mathbb{T}^3$. Given a positive function $u > 0$, consider the metric $g = u^{4} g_0$. Then we have the identity
$$ - 8 \Delta_0 u = S_g u^5 \tag{1} $$
where $S_g$ is the scalar curvature of the metric $g$. 

Note that the volume form of $g$ is $\mathrm{dvol}_g = u^6~ \mathrm{dvol}_0$. Multiply both sides of (1) by $u ~\mathrm{dvol}_0$ we find
$$ - 8 \Delta_0 u \cdot u ~\mathrm{dvol}_0 = S_g ~\mathrm{dvol}_g$$
Integrating both sides, for any non-constant $u$ you have that the left hand side is manifestly positive. But your "result" would imply that the right hand side must be non-positive. 

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More generally: 

Let $g_0$ be any constant scalar curvature ($S_0$) metric, on an $n$-dimensional manifold $M$ with $n > 2$. Let $g_u = u^{4/(n-2)} g_0$, where $u$ is a positive function. Then we have that the scalar curvature $S_u$ of $g_u$ satisfies

$$ - \gamma \Delta_0 u + S_0 u = S_u u^{2n/(n-2)} u^{-1} $$ 

where $\gamma = 4(n-1)/(n-2)$. 
Using again that $\mathrm{dvol}_u = u^{2n/(n-2)} ~\mathrm{dvol}_0$ we find that

$$ \int (- \gamma \Delta_0 + S_0)u \cdot u ~\mathrm{dvol}_0 = \int S_u ~\mathrm{dvol}_u $$

Using that $\Delta_0^{-1}$ is compact, the operator $(-\gamma \Delta_0 + S_0)$ has arbitrarily large and positive eigenvalues. This shows that you can **always** find a metric conformal to $g_0$ with positive Einstein-Hilbert integral. 

(Remark: that $g_0$ has constant scalar curvature is inessential; it just makes the description of the spectrum of $-\gamma \Delta_0 + S_0$ easier to state. You can do the same argument with any metric; or you can bring out the big guns and use Yamabe to first transform the metric to one with constant scalar curvature.)

When $n = 2$, your result is true by Gauss-Bonnet.