I do not know if there is a way to get the isoperimetric inequality from the spectral gap, but both can be proven in almost the same way. The classical references for the linear isoperimetric inequality are S.-T. Yau, "Isoperimetric constants and the first eigenvalue of a compact Riemannian manifold", Ann. Sci. École Norm. Sup. (4) 8 (1975), no. 4, 487–507 and
Yurii D. Burago and Victor A. Zalgaller, "Geometric inequalities".
I like this proof so let me give it here (this is Burago-Zalgaller presentation). For any unit tangent vector $u$ and positive real $r$, let $s(u,r)$ be the "candle function" defined by
$$dy = s(u,r) \,du \,dr$$
when $y=\exp_x(ru)$ and $u\in UT_xM$. Up to a normalization, this is simply the jacobian of the exponential map. The curvature hypothesis implies $(\log s(u,r))'\geqslant \sqrt{-\kappa}(n-1)$ where the prime denotes derivative with respect to $r$ (this is a consequence of Günther's inequality).
$\Omega$ is contained in the union of all geodesic rays from any fixed point $x_0$ to $\partial \Omega$. Let $U\subset UT_{x_0}M$ be the set of unit vectors generating geodesics that intersect $\Omega$, and for $u\in U$ let $r_u$ be the last intersection time of the geodesic generated by $u$ with $\Omega$. Then
$$\mathrm{Vol}(\partial \Omega) \geqslant \int_U s(u,r_u) \,du$$
and
$$\mathrm{Vol}(\Omega) \leqslant \int_U \int_0^{r_u} s(u,t) \,dt\,du.$$
Now, writing $s(u,r_u)=\int_0^{r_u} s'(u,t) \,dt$ and using Günther's inequality, the desired result comes.
I cannot help self-advertising: in fact, the same conclusions (Günther inequality, hence both the linear isoperimetric inequality and the spectral gap of MCKean) hold under a weaker curvature bound (some higher but non-positive sectional curvature can be compensated by enough more negative sectional curvature in other directions). This is explained in an arXiv paper with Greg Kuperberg, "A refinement of Günther's candle inequality" [arXiv:1204.3943].
I guess Gromov wanted to say that there is a lower bound for $\mathop{\rm vol}\partial V_0$
in terms of $\mathop{\rm vol} V_0/\mathop{\rm Vol} V$, $\mathop{\rm diam}V$ and lower bound for Ricci curvature. The same proof as in "Paul Levy's Isoperimetric Inequality", gives such a bound, but it is not longer sharp.
BTW, there is an analog of Levy--Gromov for open manifolds with $\mathop{\rm Ricc}\ge 0$.
It is sharp and gives a lower bound for $\mathop{\rm vol}\partial V_0$ in terms of $\mathop{\rm vol} V_0$ and the volume growth of $V$,
BUT as far as I know it is not written.
(Please correct me if I am wrong.)
Best Answer
This is definitely false. In dimension 3 if $\lambda_1,\lambda_2,\lambda_3$ are eigenvalues of the curvature operator then Ricci curvatures of eigenvectors are $\lambda_1+\lambda_2, \lambda_1+\lambda_3, \lambda_2+\lambda_3$. If one of $\lambda_i$'s is very negative but the other two are very positive then all these sums can be bounded below. Every algebraic curvature tensor can be realized at a point so locally this can definitely happen. Complete examples exist too. I can't seem to find a reference to an explicit example but it's well known that there are examples of metrics on $\mathbb R^3$ with $Ric\ge 0$ which have some negative sectional curvature. If you rescale such metric by a small number Ricci curvature remains nonnegative but sectional curvature can become arbitrary negative in some 2-planes.