No. Consider the 3-dimensional affine subspace consisting of all matrices having $1,1,-1$ on the diagonal (non-diagonal entries are arbitrary). A small perturbation of this subspace cannot intersect the cone of definite matrices. Indeed, a matrix from a perturbed subspace either has diagonal entries close to $1,1,-1$, or the maximum non-diagonal entry is much larger than the diagonal ones (and in this case, the respective $2\times 2$ minor is negative).

Even 4-dimensional subspace is not enough. Indeed, let $C$ denote the set of nonnegative matrices. It is a sharp (i.e. not containing straight lines) closed convex cone in the 6-dimensional space of all symmetric matrices. Therefore it is a cone over a compact convex set $K$ lying in a 5-dimensional affine subspace (e.g. in the subspace of matrices with trace 1). In this affine subspace, there is a 4-dimensional subspace $L$ which does not intersect $K$ (for example, the set of matrices with trace 1 and the upper-left entry equal to 2). Consider the distance from the set $C\cup(-C)$ as a function on $L$. It is positive and have a linear growth rate at infinity, and these properties are preserved under small perturbations of $L$.