An electric field is a vector field, which assigns a vector to each point in space. A vector itself cannot be negative or positive (unless we consider the one-dimensional case where a sign is meant to designate the direction). Arguing about the sign of the electric field vector generally makes no sense.
Unfortunately, your question is ambiguous, but I can consider three different ways of interpreting the post.
In the three-dimensional case, what your equation should be written as is
$$ \vec{E}(\vec{x}) = \frac{kQ}{r^{2}} \hat{r} $$
where $\hat{r}$ is the unit vector pointing from the charge $Q$ to the point in space $\vec{x}$.
Clearly, this does not have a sign.
Instead, it has a direction along with a magnitude (with one exception which is that if its magnitude is zero, then the direction is not well-defined).
The magnitude of a vector $\vec{v} = v_{1}\hat{x} + v_{2}\hat{y} + v_{3}\hat{z}$ is $|\vec{v}| = \sqrt{v_{1}^{2} + v_{2}^{2} + v_{3}^{2}}$ in the 3D case. So in the case of the electric field, we find
$$ |\vec{E}(\vec{x})| = \frac{k|Q|}{r^{2}}, $$
which is indeed always nonnegative.
Therefore, if you are talking about the magnitude of the electric field vector, then it must be nonnegative.
The only case where you could possibly say that the electric field is negative is if you're considering a one-dimensional scenario. In 1D, a vector only has one component and we write $\vec{v} = v_{1} \hat{x}$. Unfortunately, people tend to use a bit of abuse of notation by identifying $\vec{v}$ with the value $v_{1}$ (which is justified because this is the only value that determines $\vec{v}$ to begin with in 1D once we fix the unit basis vector $\hat{x}$). In that case, $v_{1}$ can be positive, negative, or zero. However, the interpretation of the sign is not meant to correspond to any "amount" of anything. Instead, the sign is meant to signify the direction of the vector (left or right or zero). Lastly, note that the magnitude of the vector ends up being the absolute value $|v_{1}|$ (this is becaue $|\vec{v}| = \sqrt{v_{1}^{2}} = |v_{1}|$).
Of the three cases, I doubt #3 is the one that is relevant, and equating $\vec{v}$ with $v_{1}$ is bad practice, since it could lead to confusion. There's much more to say about this, but I won't go much further on this idea in this post.
It's likely that your teacher had #2 in mind, in which case he would have been correct. However, you did not specify that you were talking about the magnitude of the electric field vector, which means the ambiguity of your question still stands.
If you are considering the electric field vector $\vec{E}(\vec{x})$, then the question of its sign makes no sense, because vector quantities in three-dimensions don't have any sign. They only have either a non-zero magnitude and a direction OR a zero magnitude and an undefined direction.
You can analogize this by visualizing an arrow that represents the vector. Clearly an arrow has a length and a direction, but the question of whether an arrow is positive or negative makes no sense. It's length, however, is a nonnegative real number.
