This is true if $r$ is odd or $r<2d$, and false otherwise. I assume, when you say a polynomial is symmetric you mean in fact it is homogeneous.

**1.** Let $r=2d$ and take $P(x)=-x_1^2x_2^2\dots x_d^2$. The term $x_1^2\dots x_d^2$ appears in every $r$th power of a linear form with a nonnegative coefficient, so their sum plus monomials in $d-1$ variables cannot get a negative coefficient at $x_1^2\dots x_d^2$. A similar reasoning works for all even $r\geq 2d$.

**2.** Now assume that $r$ is odd or $r<2d$. In the former case we will even show that each appropriate polynomial is even a sum of $r$th powers.

Firstly, notice that
$$
(-1)^rr!y_1y_2\dots y_r=\sum_{I\subset\{1,\dots,r\}}(-1)^{|I|}
\left(\sum_{i\in I}y_i\right)^r.
$$
So the monomial $y_1\dots y_r$ is a linear combination of $r$th powers. By substitution, the same holds for every $r$th degree monomial in $x_1,\dots,x_d$. So the linear hull of all $r$th powers is the whole space of all homogeneous polynomials of degree $d$.

Let $S^{d-1}$ be a unit sphere in $\mathbb R^d$, and let $\mu$ be the usual measure on it. If $r$ is odd, then
$$
\int_{S^{d-1}}(a^Tx)^r\,d\mu(a)=0,
$$
so $0$ lies in the convex hull of all the polynomials $(a^Tx)^r$. By Caratheodory's theorem, $0$ is a finite convex combination of some of them, and (due to symmetry of the sphere) each $r$th power is involved in some such combination. This yields that $0$ lies in the (relative) interior of the cone generated by $r$th powers, so this cone is the whole space. This is exactly what we claimed.

(Some time ago I was told that this Caratheodory trick appeared in a paper by Milman and someone else; sorry for not presenting the exact reference.)

If $r$ is even but $r<2d$, then the same integral is a nonzero polynomial $Q$, but it contains only monomials of even degree in each variable; so it is a linear combination of monomials in at most $d-1$ variables (and it is invariant under sphere rotations). Then the same arguments can be applied to the factor space by $\langle Q\rangle$. We will represent each polynomial as a sum of $r$th powers plus a multiple of $Q$, which is again what we need.