Wouldn't that mean that the quadratic form $x^2+y^2+z^2+2xy+2yz$ must be nonnegative definite (as it is a band restriction of the quadratic form $x^2+y^2+z^2+2xy+2yz+2zx$, which is clearly nonnegative definite), which contradicts its value at $x=1$, $y=-1$, $z=1$ ?

(Note that I replaced your "positive definite" by "nonnegative definite" - feel free to add $\epsilon\left(x^2+y^2+z^2\right)$ to the form for some $\epsilon>0$ to keep everything positive.)

EDIT: There's a bit more to this:

Let us denote by $A\ast B$ the *Hadamard product* of two $n\times n$ matrices $A$ and $B$ (defined by

$A\ast B=\left(a_{i,j}b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$,

where

$A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$

and $B=\left(b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$).

Let $A$ be a symmetric matrix. Then, (the matrix $A\ast B$ is nonnegative definite for every nonnegative definite matrix $B$) if and only if the matrix $A$ is nonnegative definite. The $\Longrightarrow$ direction is more or less trivial (just take $B$ to be the matrix $\left(1\right)_{1\leq i\leq n,\ 1\leq j\leq n}$) and disproves your conjecture (by taking $A$ to be the matrix whose $\left(i,j\right)$-th entry is $1$ if $\left|i-j\right|\leq d$ and $0$ otherwise). The $\Longleftarrow$ direction is interesting and most easily proven by decomposing the matrix $A$ in the form $u_1u_1^T+u_2u_2^T+...+u_nu_n^T$, where $u_1$, $u_2$, ..., $u_n$ are appropriate vectors. Another proof reduces it to Corollary 2 in my answer to MathOverflow #19100 - do you see how?