In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar.

For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $M_1$ which is the metric product of a ball of radius r in hyperbolic space $\mathbb{H}^2$ (with sectional curvature 1) and the sphere $\mathbb{S}^2(r)$. For the second manifold, consider the $4$-sphere $\mathbb{S}^4(s)$.

The volume of $M_1$ is $\sinh(t) \times 4 \pi r^2$ while its diameter is $\sqrt{\pi^2 r^2+ 4t^2} < \pi r +2 t$. The scalar curvature is $2(\frac{1}{r^2}-1)$. Whenever $r<1$, the scalar curvature of $M_1$ is positive. Meanwhile, the volume of the $4$-sphere is $\frac{8}{3}\pi^2 s^4$, its diameter is $\pi s$, and its scalar curvature is $\frac{12}{s^2}$.

To force our first manifold to have much larger volume then the second, we set $s$ very large so that the scalar curvature is less than $1$.
We also set $r=1/2$ and $t=s-1/2$. In this case, the scalar curvature of $M_1$ is 6 and the diameter of $M_1$ is strictly smaller than that of $\mathbb{S}^4(s)$. However, since hyperbolic sine grows exponentially in $t$ whereas the volume of the $4$-sphere grows as $s^4$, for sufficiently large $s$ the volume of $M_1$ is larger than that of the sphere. By setting $s$ arbitrarily large, we can force $M_1$ to have much larger volume than $\mathbb{S}^4$.