Let $Q^p$ be a continuous family of real quadratic forms over $\mathbb{R}^n$ with respect to a parameter $p$.
The hypothesis that we have is that for every $v\in \mathbb{R}^n$ there exists a $q$ such that $Q^q(v) \leq 0$.
The goal is to prove that there exists a q and an orthonormal basis $(e_i)_{1\leq i\leq n}$ of $\mathbb{R}^n$ such that $\sum_{i=1}^{n}Q^q(e_i)\leq 0$.
Is that possible to prove ?
Thank you for your time, any help is much appreciated.
The claim is false.
Consider the quadratic form $$Q^{I_2}(x,y)=x^2-y^2$$ and extend $Q$ to a continuous family parameterized by $P\in\mathrm{SO}(\mathbb{R}^2)$ via $$Q^P(\vec{v})=Q^{I_2}(P\vec{v})$$ Certainly, $Q^{I_2}(0,y)\leq0$. Since each vector $\vec{v}$ can be rotated until it points along the $y$-axis, there exists $P$ for which $Q^P(\vec{v})\leq0$. Thus $\{Q^P\}_{P\in\mathrm{SO}(\mathbb{R}^2)}$ satisfies your hypotheses.
But now fix $q$ and consider $Q^q$. Rotation preserves orthonormal bases, so we may assume $q=I_2$ w/oLoG. Any orthonormal basis of $\mathbb{R}^2$ is of the form $((x,y),(-y,x))$, where $x^2+y^2=1$. Thus we should be able to find some $(x,y)$ with \begin{gather*} x^2+y^2=1 \\ x^2-y^2\leq0 \\ y^2-x^2\leq0 \end{gather*} But it is easy to verify that these three conditions are incompatible, and so your conclusion fails.