Isn't it always possible that one can perturb the metric such that it has positive sectional curvature at a point? This is the famous theorem of Gromoll and Meyer:

Theorem (Gromoll-Meyer, 1974) There is an exotic 7-sphere with nonnegative
  sectional curvature and positive sectional curvature at a point.

I don't understand the second part of theorem "positive sectional curvature at a point". Isn't it always possible that one cane perturb the metric such that it has positive sectional curvature at a point and we have $\sec_{\min}(M)\leq \sec_q\leq \sec_p$ for all $q$ in a small neighborhoods of $p$?
 A: The short answer is that you cannot necessarily perturb a metric to get positive sectional curvature while still maintaining non-negative/positive sectional curvature everywhere else.
My favorite example of this is due to Wilking

 Manifolds with positive sectional curvature almost everywhere, Inventiones mathematicae 148(1):117-141, 2002

He constructs a smooth Riemannian metric $g$ on $M:=\mathbb{R}P^2\times \mathbb{R}P^3$ with the following properties.
1)  It is non-negatively curved everywhere.
2)  If $U\subseteq M$ is the subset of all points $p$ for which every $2$-plane $\sigma \subseteq T_p M$ is positively curved, then $U$ has full measure in $M$
In other words, in the measure-theoretic sense, $M$ is positively curved almost everywhere (and non-negatively curved everywhere).
Note, however, that $M$ is non-orientable.  The classical Synge theorem says that in odd dimensions, a positively curved closed Riemannian manifold must be orientable.  Thus, the metric $g$ cannot be deformed to being positively curved everywhere, despite the fact that it started off with positive curvature almost everywhere.
Edit  Following Travis's suggestion, I'll move part of my comment below into the answer here.
Proposition.  Suppose $(M,g)$ is a Riemannian manifold and $p\in M$.  Let $p\in U\subseteq M$ be an open set diffeomorphic to a ball in $\mathbb{R}^n$  Then $g$ can be deformed to a metric $g_1$ for which all sectional curvatures are positive near $p$, but $g= g_1$ outside of $U$.
Proof:  Because $U$ is diffeomorphic to a ball, there is a diffeomorphism $f:U\rightarrow D^n_+\subseteq S^n$, where $D^n_+$ denotes the open northern hemisphere.  Let $g_0$ denote the canonical (positively curved) metric on $S^n$.
Choose an open set $V\subseteq U$ with $\overline{V}\subseteq U$ and let $\lambda:M\rightarrow \mathbb{R}$ be a bump function with $\lambda \equiv 1$ near $p$, and $\operatorname{supp} \lambda \subseteq \overline{V}$.
The family of metrics $(1-t)g + \lambda t f^\ast g_0$ has the required properties.
A: 
Isn't it always possible that one can perturb the metric such that it has positive sectional curvature at a point and we have "$\sec_{\min}(M)\leq \sec_q\leq \sec_p$" for all $q$ in a small neighborhoods of $p$?

This statement isn't quite precise, since $\sec_a$ is not a scalar but rather a map $$Gr(2, T_a M) \to \Bbb R .$$ In the special case that $(M, g)$ is a surface (i.e., $\dim M = 2$), however, $Gr(2, T_a M)$ is the singleton set $\{T_a M\}$, so we may identify $\operatorname{sec}_a$ with a scalar and hence view $\operatorname{sec}$ as a function $M \to \Bbb R$, namely, the Gaussian curvature, $K$, and hence interpret the inequalities in the quotation in the usual way.
A standard partition-of-unity argument shows that for any smooth surface $(M, g)$ and point $p \in M$ one can deform $g$ to $g'$ in some neighborhood of $p$ so that for all $q$ in some (a priori possibly smaller) neighborhood we have $\inf K(g) \leq K'_q \leq K'_p$, where $K'$ is the Gaussian curvature of $g'$. (See Jason DeVito's good answer for details of this argument.)
This result, however, is manifestly local, and it does not guarantee the global conclusion that $K'_a \geq \inf K(M)$ for all $a \in M$.

Example Consider the flat torus $(S^1 \times S^1, g)$, for which $K \equiv 0$; in particular $g$ has nonnegative sectional curvature. We can perturb $g$ to produce a metric $g'$ with positive Gaussian curvature $K'_p$ at a point $p$. But the Gauss-Bonnet Theorem gives that
  $$\int_{S^1 \times S^1} K' \,dA = 2 \pi \chi(S^1 \times S^1) = 0.$$ By continuity $K'$ is positive in some neighborhood of $p$, which thus gives a positive contribution to the integral. To satisfy the above equality, $K'$ must also be negative on some open subset of $M$.

In short, the existence of a metric on a given manifold with nonnegative sectional curvature does not imply the existence of a metric on the same manifold with nonnegative sectional curvature at a point and positive sectional curvature at some point.
