I have a more complete answer to offer although there is still a lot that I don't know. I can give the complete story for the case $k=2$ (a pair of quadratic forms). At the end I will provide two references where details can be found. First, let's begin with the following observation that applies to arbitrary $k \ge 2$.
Let $K$ be an infinite field with characteristic different from $2$ and let $Q_1, \ldots, Q_k \in K[x_1, \ldots, x_n]$ be quadratic forms. There is an associated $n \times n$ symmetric matrix $M_i$ associated to $Q_i$,
$1 \le i \le k$.
Let $f(\lambda_1, \ldots, \lambda_k) =
\det(\lambda_1 M_1 + \cdots + \lambda_n M_n)$.
Then there exists a quadratic form $\sum_{i=1}^k \lambda_i Q_i$ with each $\lambda_i \in K$, not all zero, such that $\sum_{i=1}^k \lambda_i Q_i$ has rank $n$ if and only if $f$ is a nonzero polynomial. (This is where we need that $K$ is an infinite field.)
The question is now whether $f$ can be the zero polynomial. The previous comments show that this can indeed occur. When $k = 2$, we can give a complete characterization when this occurs. We begin with a set of basic examples.
Let $m \ge 1$ be an integer.
Let $q_1 = x_1 x_2 + x_3 x_4 + \cdots + x_{2m -1}x_{2m}$ and $q_2 = x_2 x_3 + x_4x_5 + \cdots + x_{2m} x_{2m+1}$.
Then $\lambda_1 q_1 + \lambda_2 q_2 = x_2(\lambda_1 x_1 + \lambda_2 x_3) + \cdots + x_{2m}(\lambda_1 x_{2m-1} + \lambda_2 x_{2m+1})$. It follows that every nontrivial linear combination of $q_1, q_2$ has rank $2m$, and thus
no nontrivial linear combination of $q_1, q_2$ has full rank $2m+1$.
Thus for each $m \ge 1$, we have a pair of quadratic forms $q_1, q_2$ in $2m+1$ variables with the property that no nontrivial linear combination of $q_1, q_2$ has full rank $2m + 1$. We can construct new examples by taking orthogonal direct sums of these examples in disjoint variables. Let $P_1, P_2$ be a pair of quadratic forms obtained in this way. We can write
$P_1 = q_1 \perp q_1' \perp \cdots \perp q_1^{''}$ and
$P_2 = q_2 \perp q_2' \perp \cdots \perp q_2^{''}$. Then
$\det(\lambda_1 P_1 + \lambda_2 P_2)$ is the zero polynomial.
Now let $R_1, R_2$ be any pair of quadratic forms with the property that $\det(\lambda_1 R_1 + \lambda_2 R_2)$ is a nonzero polynomial.
Finally, let $Q_1 = P_1 \perp R_1$, $Q_2 = P_2 \perp R_2$.
Then $\det(\lambda_1 Q_1 + \lambda_2 Q_2)$ is the zero polynomial.
The big theorem is that the converse is true:
Theorem. Let $K$ be an infinite field and assume that $K$ has characteristic different from $2$. Let $Q_1, Q_2 \in K[x_1, \ldots, x_n]$ be quadratic forms
and assume that $\det(\lambda Q_1 + \mu Q_2)$ is the zero polynomial in $\lambda_1, \lambda_2$. Then $Q_1, Q_2$ is given by the above construction for appropriate pairs $(q_1,q_2), (q_1', q_2'), \ldots, (q_1^{''}, q_2^{''})$, and a pair $(R_1, R_2)$. Moreover, the values of the various $m$'s used above are uniquely determined, and the pair $(R_1, R_2)$ is uniquely determined up to isomorphism (defined in a suitable manner).
A version of this result also holds over finite fields but I have left out this version to avoid even more technical details.
These results go back to Weierstrass, Kronecker, and Dickson in some form.
A modern treatment was given in
William Waterhouse,
Pairs of quadratic forms, Invent. Math. 37(1976), no.2, 157–164.
An exposition of the same material with far more details was given in
David Leep, Laura Schueller,
Classification of pairs of symmetric and alternating bilinear forms. Exposition. Math.17(1999), no.5, 385–414.
The case $k \ge 3$ is wide open as far as I know. The previous comments gave an example where every nontrivial linear combination of quadratic forms has rank $2$. I would guess that much more work could still be done when $k \ge 3$.
